Range of a Function

The range of a function \( f(x) \) is the set of all possible values that \( f(x) \) can take as \( x \) varies over its domain. Formally, if the domain of \( f(x) \) is \( D \subseteq \mathbb{R} \), then the range is given by:

\[ \text{Range of } f(x) = \{y \in \mathbb{R} : y = f(x) \text{ for some } x \in D\}. \]

It represents the actual output values of the function, considering all inputs from the domain.

The range is directly affected by the domain of the function. If the domain changes, it may result in a change in the range because the set of inputs determines the outputs. Therefore, before calculating the range, we must carefully determine the domain of the function.

Finding the range of a function involves determining the set of all possible outputs \( y \) that the function \( f(x) \) produces for a given domain. Unlike domain calculation, which often involves straightforward restrictions on the input, determining the range requires analyzing the behavior of the function and applying various techniques.

For a given domain, we explore what values \( f(x) \) can achieve by examining the function's structure, solving equations, considering limits, or applying transformations. Since different types of functions behave differently, finding the range is often a more nuanced and challenging process compared to domain calculation.

Let us start with this theorem, which can be applied quite frequently.

Theorem: Range of the Sum or Difference of Two Functions

Let there be two functions \( f(x) \) and \( g(x) \) having domains \( D_1 \) and \( D_2 \). In the domain \( D_1 \cap D_2 \), if the range of \( f(x) \) is \( \mathbb{R} \) and the range of \( g(x) \) is a bounded interval, then the range of \( f \pm g \) is also \( \mathbb{R} \).

Example:

For \( f(x) = x^3 \) and \( g(x) = \sin x \), the range of \( f(x) + g(x) \) is \( \mathbb{R} \), since the range of \( x^3 \) is \( \mathbb{R} \) and the range of \( \sin x \) is the bounded interval \([-1, 1]\).
For \( f(x) = \tan x \) and \( g(x) = \tan^{-1} x \), the range of \( f(x) + g(x) \) is \( \mathbb{R} \), as \( \tan x \) has range \( \mathbb{R} \) and \( \tan^{-1} x \) is bounded.

Example

Find the range of \( f(x) = x^3 - 100x + \sin x \), \(x \in \mathbb{R}\)

Solution:

The given function \( f(x) \) is the sum of two terms:

\( x^3 - 100x \), an odd-degree polynomial. Since odd-degree polynomials are unbounded and continuous, the range of \( x^3 - 100x \) is \( \mathbb{R} \).
\( \sin x \), whose range is the bounded interval \([-1, 1]\).

The sum of these two functions, \( f(x) = (x^3 - 100x) + \sin x \), combines a term with range \( \mathbb{R} \) and a term with range \([-1, 1]\). Adding a bounded term (like \( \sin x \)) to an unbounded function does not restrict the overall range, as the unbounded term dominates. Therefore, \( f(x) \) can take any real value.

The range of \( f(x) \) is:

\[ \boxed{\mathbb{R}} \]

Using the Graphs to Find the Range

We employ the graph method to determine the range when the graph of the function can be easily constructed. This is feasible in cases such as:

When the function can be expressed as a transformation (e.g., shifts, stretches, or reflections) of a well-known function, making it straightforward to deduce the graph.
When the graph is inherently simple, such as for linear, quadratic, or basic trigonometric functions.
When the function is defined piecewise, and each segment of the function can be expressed as a simple mathematical expression.

By constructing the graph through these methods and analyzing its behavior, the range can be directly observed from the output values covered by the graph. This visual approach simplifies the process and provides clarity in identifying the function's range.

Example

Find the Range of \( f(x) = |x-1| + |2x-3| \)

To find the range, we break the function into intervals based on the critical points where the absolute values change behavior and then define the function piecewise.

Step 1: Analyze the Absolute Values

The function \( f(x) = |x-1| + |2x-3| \) involves two absolute values:

\( |x-1| \), which changes sign at \( x=1 \):
- For \( x \geq 1 \), \( |x-1| = x-1 \).
- For \( x < 1 \), \( |x-1| = -(x-1) = 1-x \).
\( |2x-3| \), which changes sign at \( x = \frac{3}{2} \):
- For \( x \geq \frac{3}{2} \), \( |2x-3| = 2x-3 \).
- For \( x < \frac{3}{2} \), \( |2x-3| = -(2x-3) = 3-2x \).

The critical points (\( x=1 \) and \( x=\frac{3}{2} \)) divide the real line into three intervals: \( x < 1 \), \( 1 \leq x < \frac{3}{2} \), and \( x \geq \frac{3}{2} \).

Step 2: Define \( f(x) \) in Each Interval

We compute \( f(x) \) explicitly in each interval:

Case 1: \( x \geq \frac{3}{2} \):

Both \( x-1 \) and \( 2x-3 \) are positive:

\[ f(x) = (x-1) + (2x-3) = 3x - 4. \]
Case 2: \( 1 \leq x < \frac{3}{2} \):

\( x-1 \) is positive, and \( 2x-3 \) is negative:

\[ f(x) = (x-1) - (2x-3) = -x + 2. \]
Case 3: \( x < 1 \):

Both \( x-1 \) and \( 2x-3 \) are negative:

\[ f(x) = -(x-1) - (2x-3) = -3x + 4. \]

Step 3: Graph the Function

Using the piecewise definitions:

For \( x \geq \frac{3}{2} \), \( f(x) = 3x-4 \) is a line with slope 3.
For \( 1 \leq x < \frac{3}{2} \), \( f(x) = -x+2 \) is a line with slope \(-1\).
For \( x < 1 \), \( f(x) = -3x+4 \) is a line with slope \(-3\).

Step 4: Determine the Range

From the graph, \( f(x) \) achieves its minimum value at \( x=\frac{3}{2} \), where \( f(\frac{3}{2}) = \frac{1}{2} \). As \( x \to \infty \) or \( x \to -\infty \), \( f(x) \to \infty \). Hence, the range of \( f(x) \) is:

\[ \boxed{\left[\frac{1}{2}, \infty\right)}. \]

Example

Find the Range of \( f(x) = |x+2| + |x^2 - 1| \)

Solution:

The domain of \( f(x) \) is \( \mathbb{R} \), as absolute values are defined for all real numbers. To determine the range, we analyze the function by breaking it into intervals based on the expressions inside the absolute values.

The first absolute value, \( |x+2| \), changes sign at \( x = -2 \), while the second absolute value, \( |x^2 - 1| \), changes sign at \( x = -1 \) and \( x = 1 \). These critical points divide the real line into four intervals: \( x < -2 \), \( -2 \leq x < -1 \), \( -1 \leq x < 1 \), and \( x \geq 1 \).

Using these intervals, we express \( f(x) \) piecewise:

For \( x < -2 \):

Here, \( x+2 < 0 \) and \( x^2 - 1 \geq 0 \), so:

\[ f(x) = -(x+2) + (x^2 - 1) = x^2 - x - 3. \]
For \( -2 \leq x < -1 \):

Here, \( x+2 \geq 0 \) and \( x^2 - 1 \geq 0 \), so:

\[ f(x) = (x+2) + (x^2 - 1) = x^2 + x + 1. \]
For \( -1 \leq x < 1 \):

Here, \( x+2 \geq 0 \) and \( x^2 - 1 < 0 \), so:

\[ f(x) = (x+2) - (x^2 - 1) = -x^2 + x + 3. \]
For \( x \geq 1 \):

Here, \( x+2 \geq 0 \) and \( x^2 - 1 \geq 0 \), so:

\[ f(x) = (x+2) + (x^2 - 1) = x^2 + x + 1. \]

Simplified Piecewise Definition:

\[ f(x) = \begin{cases} x^2 - x - 3 & \text{if } x < -2, \\ x^2 + x + 1 & \text{if } -2 \leq x < -1, \\ -x^2 + x + 3 & \text{if } -1 \leq x < 1, \\ x^2 + x + 1 & \text{if } x \geq 1. \end{cases} \]

Step 2: Analyze the Range

From the piecewise definition, we observe:

For \( x < -2 \), \( f(x) = x^2 - x - 3 \) is a parabola opening upwards.
For \( -2 \leq x < -1 \), \( f(x) = x^2 + x + 1 \) is a parabola opening upwards.
For \( -1 \leq x < 1 \), \( f(x) = -x^2 + x + 3 \) is a parabola opening downwards.
For \( x \geq 1 \), \( f(x) = x^2 + x + 1 \) continues as a parabola opening upwards.

The minimum value occurs at \( x = -1 \), where \( y = 1 \). The function grows unbounded as \( x \to \pm\infty \). Hence, the range of \( f(x) \) is:

\[ \boxed{[1, \infty)}. \]

Periodic Functions

For a periodic function, the range can be determined by finding the set of output values within a single fundamental period. Since the behavior of a periodic function repeats identically over each period, the range calculated in one period applies to the entire function.

Continuous Monotonic Functions

For a continuous monotonic function \( f(x) \), the range can be found by evaluating the function at the endpoints of its domain or taking the limits at these endpoints.

For an increasing function \( f(x) \) defined on \([a, b]\), the function's output values increase as \( x \) moves from \( a \) to \( b \). Thus, the range is:

\[ [f(a), f(b)]. \]
For a decreasing function \( f(x) \) defined on \([a, b]\), the function's output values decrease as \( x \) moves from \( a \) to \( b \). Therefore, the range is:

\[ [f(b), f(a)]. \]

What is a continuous monotonic fucntion?

Even though we shall study in detail about this in later chapters, it would be good that you should have some basic idea about it.

A continuous function is one where the graph of the function has no breaks, jumps, or holes. In simpler terms, as you move along the graph, you can trace it without lifting your pen. For example, if you plot the function \( f(x) = x^2 \), the curve will be smooth and continuous, with no interruptions.

Now, a monotonic function is a function that either always increases or always decreases. This means that if you pick two values of \( x \) from the graph, the function will either always give larger outputs as \( x \) increases (if the function is increasing), or always give smaller outputs as \( x \) increases (if the function is decreasing).

A monotonically increasing function means that for any two numbers \( x_1 \) and \( x_2 \) where \( x_1 < x_2 \), the function values satisfy \( f(x_1) \leq f(x_2) \). This means that as \( x \) increases, the function's value doesn’t decrease.
A monotonically decreasing function means that for any two numbers \( x_1 \) and \( x_2 \) where \( x_1 < x_2 \), the function values satisfy \( f(x_1) \geq f(x_2) \). This means that as \( x \) increases, the function's value doesn’t increase.

When we say that a function is continuous and monotonic, it means that the function’s graph is smooth (no jumps or breaks) and that it keeps increasing or decreasing steadily without changing direction.

Consider the function \( f(x) = 3x \): - This function is continuous because you can draw its graph without lifting your pen. - It is also monotonic because as \( x \) increases, \( 3x \) keeps increasing steadily.

In contrast, a function like \( f(x) = x^2 \) is continuous but not monotonic because it increases for \( x > 0 \) and decreases for \( x < 0 \).

Composition of Functions with Continuous Monotonic Outer Function

In a composition \( f(g(x)) \), if \( f \) is a continuous monotonic function (strictly increasing or decreasing) over the range of \( g(x) \) for the given domain, then the range of \( f(g(x)) \) can be determined by finding the range of \( g(x) \) and applying \( f \) to the endpoints of that range.

Example: Find the Range of \( f(x) = e^{x^3 + 2x - 3} \) in the Domain \( [-1, 1] \)

Solution:

The function \( e^x \) is continuous and strictly increasing for all \( x \). To determine the range of \( f(x) \), we first find the range of \( g(x) = x^3 + 2x - 3 \) in the domain \( [-1, 1] \).

From the previous problem, we know that the range of \( g(x) \) in \( [-1, 1] \) is \( [6, 0] \). Since \( e^x \) is strictly increasing, applying it to \( [6, 0] \) gives:

\[ \text{Range of } f(x) = \left[e^6, e^0\right] = \left[e^6, 1\right]. \]

Thus,

the range of \( f(x) = e^{x^3 + 2x - 3} \) in \( [-1, 1] \) is:

\[ \boxed{[e^6, 1]}. \]

Using Algebraic and Trigonometric Techniques to Find the Range

(a) AM-GM Inequality

One of the most commonly used inequalities for finding the range of a function is the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states:

\[ \frac{a+b}{2} \geq \sqrt{ab}, \quad \text{for non-negative numbers } a \text{ and } b. \]

Equality occurs if and only if \( a = b \). The generalized form for \( n \) non-negative terms \( a_1, a_2, \dots, a_n \) is:

\[ \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \dots a_n}, \]

where equality occurs if and only if \( a_1 = a_2 = \dots = a_n \).

The AM-GM inequality is particularly useful when all terms in the expression are non-negative in the given domain and their product is constant.

Example 1: Find the Range of \( f(x) = \tan^2 x + \cot^2 x \)

Solution:

Both \( \tan^2 x \) and \( \cot^2 x \) are positive in their domains, and their product is a constant:

\[ \tan^2 x \cdot \cot^2 x = 1. \]

Applying the AM-GM inequality:

\[ \frac{\tan^2 x + \cot^2 x}{2} \geq \sqrt{\tan^2 x \cdot \cot^2 x}. \]

Simplify:

\[ \frac{\tan^2 x + \cot^2 x}{2} \geq \sqrt{1}. \]

\[ \tan^2 x + \cot^2 x \geq 2. \]

Since \( \tan^2 x + \cot^2 x \) can take arbitrarily large values as \( x \to \pm \frac{\pi}{2} \), the range is:

\[ \boxed{[2, \infty)}. \]

Example 2: Find the Range of \( f(x) = x^6 + \frac{6}{x^2} \) in its Exhaustive Domain

Solution:

We write \( f(x) = x^6 + \frac{6}{x^2} \). To apply AM-GM, we split \( \frac{6}{x^2} \) into three equal parts:

\[ f(x) = x^6 + \frac{2}{x^2} + \frac{2}{x^2} + \frac{2}{x^2}. \]

Now, applying the AM-GM inequality for four terms:

\[ \frac{x^6 + \frac{2}{x^2} + \frac{2}{x^2} + \frac{2}{x^2}}{4} \geq \sqrt[4]{x^6 \cdot \frac{2}{x^2} \cdot \frac{2}{x^2} \cdot \frac{2}{x^2}}. \]

Simplify the product:

\[ \frac{f(x)}{4} \geq \sqrt[4]{x^6 \cdot \frac{8}{x^6}} = \sqrt[4]{8}. \]

\[ f(x) \geq 4 \cdot \sqrt[4]{8} = 4 \cdot 2^{3/4}. \]

Thus, the minimum value of \( f(x) \) is \( 4 \cdot 2^{3/4} = 2^{11/4} \). Since \( x^6 \) can grow arbitrarily large, the range is:

\[ \boxed{\left[2^{11/4}, \infty\right)}. \]

(b) Using the Completing the Square Method for Quadratic Expressions

Completing the square is a frequently used method to find the range of functions that can be expressed as quadratic expressions. A quadratic expression of the form:

\[ ax^2 + bx + c \]

can be rewritten as:

\[ a\left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a}, \quad \text{where } \Delta = b^2 - 4ac. \]

This transformation helps identify the minimum or maximum value of the quadratic expression, depending on whether it opens upwards or downwards.

Example: Find the Range of \( f(x) = 4\cos^2 x - 3\cos x + 4 \) for \( x \in \mathbb{R} \)

Solution:

Rewrite \( f(x) \) in terms of \( \cos x \) using the completing the square method:

\[ f(x) = 4\cos^2 x - 3\cos x + 4. \]

Factorize the quadratic expression in \( \cos x \):

\[ f(x) = 4\left(\cos^2 x - \frac{3}{4}\cos x + 1\right). \]

Complete the square for \( \cos^2 x - \frac{3}{4}\cos x \):

\[ \cos^2 x - \frac{3}{4}\cos x = \left(\cos x - \frac{3}{8}\right)^2 - \frac{9}{64}. \]

Substitute this back:

\[ f(x) = 4\left(\left(\cos x - \frac{3}{8}\right)^2 - \frac{9}{64} + 1\right). \]

Simplify:

\[ f(x) = 4\left(\left(\cos x - \frac{3}{8}\right)^2 + \frac{55}{64}\right). \]

Distribute \( 4 \):

\[ f(x) = 4\left(\cos x - \frac{3}{8}\right)^2 + \frac{220}{64} = 4\left(\cos x - \frac{3}{8}\right)^2 + \frac{55}{16}. \]

Now, for \( x \in \mathbb{R} \), \( \cos x \in [-1, 1] \). Hence:

\[ \cos x - \frac{3}{8} \in \left[-1 - \frac{3}{8}, 1 - \frac{3}{8}\right] = \left[-\frac{11}{8}, \frac{5}{8}\right]. \]

The square term satisfies:

\[ 0 \leq \left(\cos x - \frac{3}{8}\right)^2 \leq \left(\frac{11}{8}\right)^2 = \frac{121}{64}. \]

Substitute back into \( f(x) \):

\[ f(x) = 4\left(\cos x - \frac{3}{8}\right)^2 + \frac{55}{16}. \]

We know that \( \cos x \) takes values in the range:

\[ -1 \leq \cos x \leq 1 \]

We now look at \( \cos x - \frac{3}{8} \). Subtracting \( \frac{3}{8} \) from both sides of the inequality:

\[ -1 - \frac{3}{8} \leq \cos x - \frac{3}{8} \leq 1 - \frac{3}{8} \]

This simplifies to:

\[ -\frac{11}{8} \leq \cos x - \frac{3}{8} \leq \frac{5}{8} \]

Now, we square both sides of the inequality:

\[ 0 \leq \left( \cos x - \frac{3}{8} \right)^2 \leq \left( \frac{11}{8} \right)^2 = \frac{121}{64} \]

Next, we multiply the entire inequality by 4:

\[ 0 \leq 4 \left( \cos x - \frac{3}{8} \right)^2 \leq 4 \times \frac{121}{64} = \frac{484}{64} = \frac{121}{16} \]

Finally, we add \( \frac{55}{16} \) to the entire inequality:

\[ 0 + \frac{55}{16} \leq 4 \left( \cos x - \frac{3}{8} \right)^2 + \frac{55}{16} \leq \frac{121}{16} + \frac{55}{16} \]

Simplifying the bounds:

\[ \frac{55}{16} \leq f(x) \leq \frac{176}{16} = 11 \]

Thus, the range of \( f(x) \) is:

\[ \left[ \frac{55}{16}, 11 \right] \]

\(\blacksquare\)

Example

Find the range of the function \( f(x) = 3e^{2x} - e^x + 1 \) where \( x \in \mathbb{R} \).

Solution:

We are given the function:

\[ f(x) = 3e^{2x} - e^x + 1 \]

Let \( t = e^x \), so that \( t > 0 \) for all \( x \in \mathbb{R} \). The function \( f(x) \) becomes:

\[ f(x) = 3e^{2x} - e^x + 1 = 3t^2 - t + 1 \]

Thus, we need to find the range of the function:

\[ g(t) = 3t^2 - t + 1 \]

where \( t \in (0, \infty) \).

To analyze the range of \( g(t) \), we begin by completing the square. First, factor out the 3 from the first two terms:

\[ g(t) = 3\left(t^2 - \frac{1}{3}t\right) + 1 \]

Next, complete the square inside the parentheses. We get,

\[ g(t) = 3\left(\left(t - \frac{1}{6}\right)^2 - \frac{1}{36}\right) + 1 \]

Simplify:

\[ g(t) = 3\left(t - \frac{1}{6}\right)^2 - \frac{3}{36} + 1 \]

\[ g(t) = 3\left(t - \frac{1}{6}\right)^2 - \frac{1}{12} + 1 \]

\[ g(t) = 3\left(t - \frac{1}{6}\right)^2 + \frac{11}{12} \]

Since \( \left(t - \frac{1}{6}\right)^2 \geq 0 \) for all \( t \), the minimum value of \( 3\left(t - \frac{1}{6}\right)^2 \) is 0, which occurs when \( t = \frac{1}{6} \). As \( t \to \infty \), \( 3\left(t - \frac{1}{6}\right)^2 \to \infty \).

Thus, the range of \( g(t) = 3\left(t - \frac{1}{6}\right)^2 + \frac{11}{12} \) is:

\[ \left[ \frac{11}{12}, \infty \right) \]

Range of some common Trigonometric Expressions

I. Range of \( a \sin(kx + \varphi) + b \)

Consider the expression \( y = a \sin(kx + \varphi) + b \), where \( a \), \( k \), \( b \), and \( \varphi \) are constants, and \( x \in \mathbb{R} \). The sine function, \( \sin(kx + \varphi) \), has a range of \( [-1, 1] \). By multiplying this range by \( a \), the new range of \( a \sin(kx + \varphi) \) becomes \( [-a, a] \). Adding \( b \) shifts this range vertically, resulting in the final range of the function:

\[ [b - a, b + a] \]

II. Range of the Expression \( a \sin(kx + \varphi) + b \cos(kx + \varphi) \)

To determine the range of the expression \( a \sin(kx + \varphi) + b \cos(kx + \varphi) \), where \( a \), \( b \), \( k \), and \( \varphi \) are constants, and \( x \in \mathbb{R} \), we proceed as follows:

First, we factor out \( \sqrt{a^2 + b^2} \) from the terms involving sine and cosine:

\[ a \sin(kx + \varphi) + b \cos(kx + \varphi) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin(kx + \varphi) + \frac{b}{\sqrt{a^2 + b^2}} \cos(kx + \varphi) \right) \]

Next, we introduce the angle \( \theta \) defined as:

\[ \frac{a}{\sqrt{a^2 + b^2}} = \cos \theta \quad \text{and} \quad \frac{b}{\sqrt{a^2 + b^2}} = \sin \theta \]

Since \( \cos^2 \theta + \sin^2 \theta = 1 \), this assumption holds true. Substituting these into the expression, we get:

\[ a \sin(kx + \varphi) + b \cos(kx + \varphi) = \sqrt{a^2 + b^2} \left( \cos \theta \sin(kx + \varphi) + \sin \theta \cos(kx + \varphi) \right) \]

Using the sum of angles identity for sine, \( \cos \theta \sin(kx + \varphi) + \sin \theta \cos(kx + \varphi) = \sin(kx + \varphi + \theta) \), we obtain:

\[ a \sin(kx + \varphi) + b \cos(kx + \varphi) = \sqrt{a^2 + b^2} \sin(kx + \varphi + \theta) \]

Since \( \sin(kx + \varphi + \theta) \) takes values in the range \( [-1, 1] \), the range of the expression is:

\[ \left[ -\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2} \right] \]

Thus, the range of \( a \sin(kx + \varphi) + b \cos(kx + \varphi) \) is:

\[ \boxed{[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]} \]

III. Range of the Expression \( a \cos^2 x + b \sin x \cos x + c \sin^2 x \)

To find the range of the expression \( a \cos^2 x + b \sin x \cos x + c \sin^2 x \), we begin by expressing the trigonometric functions in terms of \( \cos 2x \) and \( \sin 2x \).

We start by recalling the standard identities:

\[ \cos^2 x = \frac{1 + \cos 2x}{2}, \quad \sin^2 x = \frac{1 - \cos 2x}{2}, \quad \sin x \cos x = \frac{1}{2} \sin 2x \]

Substitute these identities into the given expression:

\[ a \cos^2 x + b \sin x \cos x + c \sin^2 x \]

\[ = a \left(\frac{1 + \cos 2x}{2}\right) + b \left(\frac{1}{2} \sin 2x \right) + c \left(\frac{1 - \cos 2x}{2}\right) \]

Now, simplify the expression:

\[ = \frac{a(1 + \cos 2x)}{2} + \frac{b \sin 2x}{2} + \frac{c(1 - \cos 2x)}{2} \]

\[ = \frac{a + c}{2} + \frac{a - c}{2} \cos 2x + \frac{b}{2} \sin 2x \]

We now have the expression in a form similar to the previous theorem. The expression is of the form \( p + q \cos 2x + r \sin 2x \), where \( p = \frac{a + c}{2} \), \( q = \frac{a - c}{2} \), and \( r = \frac{b}{2} \).

Using the theorem, the range of the expression \( p + q \cos 2x + r \sin 2x \) is:

\[ \left[ p - \sqrt{q^2 + r^2}, p + \sqrt{q^2 + r^2} \right] \]

Substituting \( p \), \( q \), and \( r \):

\[ \left[ \frac{a + c}{2} - \sqrt{\left(\frac{a - c}{2}\right)^2 + \left(\frac{b}{2}\right)^2}, \frac{a + c}{2} + \sqrt{\left(\frac{a - c}{2}\right)^2 + \left(\frac{b}{2}\right)^2} \right] \]

Thus, the range of the expression \( a \cos^2 x + b \sin x \cos x + c \sin^2 x \) is:

\[ \boxed{\left[ \frac{a + c - \sqrt{(a - c)^2 + b^2}}{2}, \frac{a + c + \sqrt{(a - c)^2 + b^2}}{2} \right]} \]

IV. Range of the Expression \( a \cos(x + \varphi_1) + b \sin(x + \varphi_2) \)

To determine the range of the expression \( a \cos(x + \varphi_1) + b \sin(x + \varphi_2) \), where \( a \), \( b \), \( \varphi_1 \), and \( \varphi_2 \) are constants, we begin by expanding the trigonometric terms using the angle addition formulas.

Expanding \( a \cos(x + \varphi_1) \) and \( b \sin(x + \varphi_2) \):

\[ a \cos(x + \varphi_1) = a (\cos x \cos \varphi_1 - \sin x \sin \varphi_1) \]

\[ b \sin(x + \varphi_2) = b (\sin x \cos \varphi_2 + \cos x \sin \varphi_2) \]

Thus, the original expression becomes:

\[ a \cos(x + \varphi_1) + b \sin(x + \varphi_2) = (a \cos \varphi_1 + b \sin \varphi_2) \cos x + (b \cos \varphi_2 - a \sin \varphi_1) \sin x \]

This is now in the form \( A \cos x + B \sin x \), where:

\[ A = a \cos \varphi_1 + b \sin \varphi_2 \]

\[ B = b \cos \varphi_2 - a \sin \varphi_1 \]

The range of the expression \( A \cos x + B \sin x \), where \( A \) and \( B \) are constants, is known to be:

\[ \left[ -\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2} \right] \]

Substituting the values of \( A \) and \( B \), we obtain:

\[ \left[ -\sqrt{(a \cos \varphi_1 + b \sin \varphi_2)^2 + (b \cos \varphi_2 - a \sin \varphi_1)^2}, \sqrt{(a \cos \varphi_1 + b \sin \varphi_2)^2 + (b \cos \varphi_2 - a \sin \varphi_1)^2} \right] \]

Simplifying the squared terms:

\[ = \left[ -\sqrt{a^2 + b^2 + 2ab \sin(\varphi_2 - \varphi_1)}, \sqrt{a^2 + b^2 + 2ab \sin(\varphi_2 - \varphi_1)} \right] \]

Therefore, the range of the expression \( a \cos(x + \varphi_1) + b \sin(x + \varphi_2) \) is:

\[ \boxed{\left[ -\sqrt{a^2 + b^2 + 2ab \sin(\varphi_2 - \varphi_1)}, \sqrt{a^2 + b^2 + 2ab \sin(\varphi_2 - \varphi_1)} \right]} \]

V. Range of the Expression \( a \tan^2 x + b \cot^2 x \)

The range of the expression \( a \tan^2 x + b \cot^2 x \), where \( a > 0 \) and \( b > 0 \), is determined using the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

For two non-negative real numbers \( p \) and \( q \), the AM-GM inequality states:

\[ \frac{p + q}{2} \geq \sqrt{pq} \]

Applying this to the terms \( p = a \tan^2 x \) and \( q = b \cot^2 x \), the inequality becomes:

\[ \frac{a \tan^2 x + b \cot^2 x}{2} \geq \sqrt{a \tan^2 x \cdot b \cot^2 x} \]

Simplifying the right-hand side:

\[ \sqrt{a \tan^2 x \cdot b \cot^2 x} = \sqrt{ab \tan^2 x \cot^2 x} = \sqrt{ab} \]

Multiplying both sides of the inequality by 2:

\[ a \tan^2 x + b \cot^2 x \geq 2 \sqrt{ab} \]

The minimum value of \( a \tan^2 x + b \cot^2 x \) is attained when \( \tan^2 x = \cot^2 x \), i.e., when \( \tan^2 x = 1 \). In this case, the value of the expression is \( 2 \sqrt{ab} \).

Since \( \tan^2 x \) and \( \cot^2 x \) can grow without bound as \( x \) varies, the maximum value of \( a \tan^2 x + b \cot^2 x \) tends to infinity. Therefore, the range of the expression is:

\[ [2 \sqrt{ab}, \infty) \]

VI. Range of the Expression \( a \sec^2 x + b \csc^2 x \)

The expression given is of the form \( a \sec^2 x + b \csc^2 x \), where \( a > 0 \) and \( b > 0 \). To find the range of this expression, we start by rewriting \( \sec^2 x \) and \( \csc^2 x \) in terms of \( \tan^2 x \) and \( \cot^2 x \), respectively.

Using the identity \( \sec^2 x = 1 + \tan^2 x \), we have:

\[ a \sec^2 x = a (1 + \tan^2 x) \]

Similarly, using the identity \( \csc^2 x = 1 + \cot^2 x \), we get:

\[ b \csc^2 x = b (1 + \cot^2 x) \]

Thus, the expression \( a \sec^2 x + b \csc^2 x \) becomes:

\[ a (1 + \tan^2 x) + b (1 + \cot^2 x) \]

Simplifying this, we get:

\[ a + a \tan^2 x + b + b \cot^2 x = a + b + a \tan^2 x + b \cot^2 x \]

Now, we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the terms \( a \tan^2 x \) and \( b \cot^2 x \). According to the AM-GM inequality, for two non-negative real numbers \( p \) and \( q \):

\[ \frac{p + q}{2} \geq \sqrt{pq} \]

Using \( p = a \tan^2 x \) and \( q = b \cot^2 x \), we apply the AM-GM inequality:

\[ \frac{a \tan^2 x + b \cot^2 x}{2} \geq \sqrt{a \tan^2 x \cdot b \cot^2 x} \]

Simplifying the right-hand side:

\[ \sqrt{a \tan^2 x \cdot b \cot^2 x} = \sqrt{ab \tan^2 x \cot^2 x} = \sqrt{ab} \]

Thus, we have:

\[ a \tan^2 x + b \cot^2 x \geq 2 \sqrt{ab} \]

Therefore, the expression \( a + b + a \tan^2 x + b \cot^2 x \) has a minimum value of \( a + b + 2 \sqrt{ab} \). As \( \tan^2 x \) and \( \cot^2 x \) can grow without bound, the maximum value tends to infinity.

Thus, the range of the expression is:

\[ [a + b + 2 \sqrt{ab}, \infty) \]

Finally, we get,

\[ \boxed{\left[ \left( \sqrt{a} + \sqrt{b} \right)^2, \infty\right)} \]

Finding the Range Using Algebraic Transformation

Sometimes, an expression may not immediately reveal its range due to the presence of multiple terms dependent on the variable. However, through algebraic manipulation, it is possible to transform such expressions into forms where only one term remains dependent on the variable. This allows for easier determination of the range.

For example, consider the expression:

\[ \sin^4 x + \cos^4 x \]

At first glance, it is difficult to directly determine the range because both \( \sin x \) and \( \cos x \) appear in the expression. However, we can manipulate the expression to make it more manageable. Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite the expression as follows:

\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x \]

Since \( \sin^2 x + \cos^2 x = 1 \), this simplifies to:

\[ \sin^4 x + \cos^4 x = 1 - 2 \sin^2 x \cos^2 x \]

Next, we apply the identity for \( \sin 2x \), which is \( \sin 2x = 2 \sin x \cos x \), so \( \sin^2 2x = 4 \sin^2 x \cos^2 x \). Using this, we can express the original equation as:

\[ \sin^4 x + \cos^4 x = 1 - \frac{\sin^2 2x}{2} \]

Now, we are left with the expression \( 1 - \frac{\sin^2 2x}{2} \), where the only term dependent on \( x \) is \( \sin^2 2x \). The range of \( \sin^2 2x \) is easily known to be \( [0, 1] \), as \( \sin^2 2x \) takes values between 0 and 1 for all \( x \).

Thus, the range of the expression \( \sin^4 x + \cos^4 x \) is:

\[ 1 - \frac{1}{2} \leq \sin^4 x + \cos^4 x \leq 1 - 0 \]

This simplifies to:

\[ \frac{1}{2} \leq \sin^4 x + \cos^4 x \leq 1 \]

Therefore, the range of \( \sin^4 x + \cos^4 x \) is:

\[ \boxed{\left[ \frac{1}{2}, 1 \right]} \]

Have a look at another example.

Example

Find the range of \( f(x) = \frac{e^x}{e^x + 1} \) in its exhaustive domain.

Solution:

First, express \( f(x) \) as:

\[ f(x) = \frac{e^x}{e^x + 1} = \frac{e^x + 1 - 1}{e^x + 1} = 1 - \frac{1}{e^x + 1} \]

Since only one term \(e^x\) is dependent of the variable \(x\) we can easily determine the range of the expression. We know that for \( x \in \mathbb{R} \), \( 0 < e^x < \infty \).

Thus, we have:

\[ 1 < e^x + 1 < \infty \quad \Rightarrow \quad 0 < \frac{1}{e^x + 1} < 1 \]

Taking the reciprocal of \( e^x + 1 \) and then subtracting from 1:

\[ -1 < -\frac{1}{e^x + 1} < 0 \quad \Rightarrow \quad 0 < 1 - \frac{1}{e^x + 1} < 1 \]

Therefore, the range of the given function is:

\[ \boxed{(0, 1)} \]

Example

Find the range of \( f(x) = \frac{2e^x - e^{-x}}{e^x + e^{-x}} \) for \( x \in \mathbb{R} \).

Solution:

To determine the range of \( f(x) \), we will manipulate the expression such that it reduces to a form where only one term depends on \( x \). This will make analyzing the behavior of the function more straightforward.

Start with the given function:

\[ f(x) = \frac{2e^x - e^{-x}}{e^x + e^{-x}} \]

To simplify the expression, multiply both the numerator and denominator by \( e^x \), which is positive for all \( x \in \mathbb{R} \). This eliminates the negative exponent:

\[ f(x) = \frac{2e^{2x} - 1}{e^{2x} + 1} \]

Next, observe that \( e^{2x} \) appears in both the numerator and denominator. To simplify further, rewrite \( 2e^{2x} - 1 \) by adding and subtracting \( 1 \):

\[ f(x) = \frac{2(e^{2x} + 1 - 1) - 1}{e^{2x} + 1} \]

Reorganize terms to separate \( e^{2x} + 1 \) in the numerator:

\[ f(x) = \frac{2(e^{2x} + 1) - 3}{e^{2x} + 1} \]

Split the expression into two parts:

\[ f(x) = 2 - \frac{3}{e^{2x} + 1} \]

For \( x \in \mathbb{R} \), the exponential term \( e^{2x} > 0 \). Adding \( 1 \), we know:

\[ 1 < e^{2x} + 1 < \infty \]

Taking the reciprocal of this inequality gives:

\[ 0 < \frac{1}{e^{2x} + 1} < 1 \]

Now multiply through by \( 3 \):

\[ 0 < \frac{3}{e^{2x} + 1} < 3 \]

Multiply though -1:

\[ -3 < -\frac{3}{e^{2x} + 1} < 0 \]

Now add 2:

\[ -1 < 2 -\frac{3}{e^{2x} + 1} < 2 \]

Thus, \( f(x) \) is constrained to the interval:

\[ f(x) \in (-1, 2) \]

The range of \( f(x) = \frac{2e^x - e^{-x}}{e^x + e^{-x}} \) is:

\[ \boxed{(-1, 2)} \]

Find Range using Inverses

Finding the range of a function using inverses relies on the fundamental relationship between inputs and outputs of a function. For a function \( y = f(x) \), \(x\) is the input, and \(y\) is the output. To determine the range of \( f(x) \), we need to identify all possible values of \(y\). By inverting the function—isolating \(x\) in terms of \(y\)—we obtain a new expression \( x = g(y) \). The key insight is that the domain of \( g(y) \) represents all possible values of \(y\) that are valid in the original function \( f(x) \). Thus, the domain of \( g(y) \) is exactly the range of \( f(x) \).

This method works because the process of inverting the function maps outputs (\(y\)) back to their corresponding inputs (\(x\)), ensuring that the values of \(y\) for which \( g(y) \) is defined are the only values \(y\) can take in the original function.

This approach is particularly powerful when isolating \(x\) in terms of \(y\) is straightforward. For example, consider the function \( y = \frac{2}{x - 1} \). To find its range:

Start with \( y = \frac{2}{x - 1} \) and solve for \(x\):

\[ y(x - 1) = 2 \implies x - 1 = \frac{2}{y} \implies x = \frac{2}{y} + 1. \]
Analyze the resulting expression \( x = g(y) = \frac{2}{y} + 1 \). Here, \(x\) is defined for all \(y \neq 0\) because division by zero is undefined. Hence, the domain of \(g(y)\) is \(y \in (-\infty, 0) \cup (0, \infty)\).

The range of \( f(x) \) is thus \( (-\infty, 0) \cup (0, \infty) \), as these are the only values \(y\) can take for which \(x = g(y)\) is well-defined.

This method demonstrates its elegance when \(x\) can be cleanly expressed in terms of \(y\), as it directly ties the range of \(f(x)\) to the domain of the inverse \(g(y)\). However, this approach only works when it is possibele to isolate \(x\). This approach will not work on the function \(f(x) = 2x + \sin x\)

Restricted Domain

First consider the problem when the domain is not restricted. The domain is exhaustive.

Problem: Find the range of \( y = \frac{x-1}{2x+1} \).

Solution:

We start with the given equation:

\[ y = \frac{x-1}{2x+1}. \]

Rewriting it to express \( x \) in terms of \( y \):

\[ y = \frac{x-1}{2x+1} \implies x = \frac{y+1}{1-2y}. \]

For this expression to be valid, the denominator \( 1-2y \neq 0 \), which implies \( y \neq \frac{1}{2} \).

Thus, the range of \( y \) is:

\[ \mathbb{R} \setminus \left\{\frac{1}{2}\right\}. \]

Now consider the problem where we have restricted the domain.

Problem: Find the range of \( y = \frac{x-1}{2x+1} \) when \( x \geq -1 \).

Solution:

Starting with the equation:

\[ y = \frac{x-1}{2x+1} \implies x = \frac{y+1}{1-2y}. \]

Now, impose the condition \( x \geq -1 \):

\[ \frac{y+1}{1-2y} \geq -1. \]

Bring \( -1 \) to the left-hand side:

\[ \frac{y+1}{1-2y} + 1 \geq 0. \]

Simplify the expression:

\[ \frac{y+1 + (1-2y)}{1-2y} \geq 0 \implies \frac{2 - y}{1-2y} \geq 0. \]

The critical points for the inequality are obtained by setting the numerator and denominator to zero:

\( 2 - y = 0 \implies y = 2 \),
\( 1 - 2y = 0 \implies y = \frac{1}{2} \).

These points divide the number line into intervals. Using the wavy curve method and considering the signs of \( 2 - y \) and \( 1 - 2y \):

For \( y < \frac{1}{2} \), the fraction \( \frac{2-y}{1-2y} > 0 \),
For \( \frac{1}{2} < y < 2 \), the fraction \( \frac{2-y}{1-2y} < 0 \),
For \( y > 2 \), the fraction \( \frac{2-y}{1-2y} > 0 \).

Thus, \( \frac{2-y}{1-2y} \geq 0 \) holds for:

\[ y \in (-\infty, \frac{1}{2}) \cup [2, \infty). \]

Therefore, the range is:

\[ (-\infty, \frac{1}{2}) \cup [2, \infty). \]

Example

Be cautious about domain changing operations like squaring and summation of logarithms.

Problem 1: Find the range of \( y = \sqrt{\frac{x-1}{x+1}} \).

Solution:

We start with the equation:

\[ y = \sqrt{\frac{x-1}{x+1}}, \]

and observe that \( y \geq 0 \) since the square root is defined only for non-negative values.

Squaring both sides:

\[ y^2 = \frac{x-1}{x+1}. \]

From this, solve for \( x \) in terms of \( y \):

\[ x = \frac{y^2 + 1}{1 - y^2}. \]

We see that \( y^2 \neq 1 \) because this would make the denominator of \( x = \frac{y^2+1}{1-y^2} \) zero. Therefore:

\[ y \neq \pm 1. \]

Finally, since \( y \geq 0 \), the range is:

\[ [0, \infty) \setminus \{1\}. \]

\[\boxed{\text{Range: } [0, \infty) \setminus \{1\}}\]

Problem 2: Find the range of \( y = \ln(x+1) - \ln(x) \).

Solution:

Start with the given equation:

\[ y = \ln(x+1) - \ln(x) \implies y = \ln\left(\frac{x+1}{x}\right). \]

This operation changes the domain with respect to the variable \( x \). In the original equation, \( x > 0 \) (this comes from the domain of logarithmic functions). However, after the transformation, the domain is influenced by the expression \( \frac{x+1}{x} \). Hence, we must analyze carefully.

Rewriting the equation:

\[ y = \ln\left(\frac{x+1}{x}\right) \implies \frac{x+1}{x} = e^y \implies x = \frac{1}{e^y - 1}. \]

From the last equation, \( x > 0 \) implies:

\[ \frac{1}{e^y - 1} > 0 \implies e^y > 1 \implies y > 0. \]

Therefore, \( y > 0 \).

The range of \( y \) is:

\[ (0, \infty). \]

Range of Rational Functions

When studying rational functions, we focus on expressions of the form:

\[ f(x) = \frac{P(x)}{Q(x)}, \]

where \( P(x) \) and \( Q(x) \) are polynomials. The range of such functions depends heavily on the degrees of \( P(x) \) and \( Q(x) \). Specifically, we analyze rational functions of the following types:

Linear over Quadratic: \( f(x) = \frac{\text{linear}}{\text{quadratic}} \),
Quadratic over Linear: \( f(x) = \frac{\text{quadratic}}{\text{linear}} \),
Quadratic over Quadratic: \( f(x) = \frac{\text{quadratic}}{\text{quadratic}} \).

In the example below, we will determine the range of \( f(x) = \frac{x^2}{x^2 + x + 1} \), a rational function of the type quadratic over quadratic.

Example

Find the Range of \( f(x) = \frac{x^2}{x^2 + x + 1} \)

Solution:

Start with the equation:

\[ y = \frac{x^2}{x^2 + x + 1}. \]

Cross-multiply to eliminate the denominator:

\[ yx^2 + yx + y = x^2. \]

Rearranging terms, we obtain:

\[ (y-1)x^2 + yx + y = 0. \tag{1} \]

This equation represents a quadratic in \( x \). However, if \( y = 1 \), the coefficient of \( x^2 \) becomes zero, making the equation linear. Therefore, we must separately consider the case \( y = 1 \).

**Case 1: \( y \neq 1 \) **

When \( y \neq 1 \), equation (1) remains quadratic in \( x \). Using the quadratic formula:

\[ x = \frac{-y \pm \sqrt{y^2 - 4(y-1)y}}{2(y-1)}. \]

For \( x \) to have real values, the discriminant \( \Delta \) must satisfy:

\[ \Delta = y^2 - 4(y-1)y \geq 0. \]

That is,

\[ y(4 - 3y) \geq 0. \]

Using the wavy curve method, the critical points are \( y = 0 \) and \( y = \frac{4}{3} \). From the wavy curve analysis, the inequality \( y(4 - 3y) \geq 0 \) holds when:

\[ y \in [0, \frac{4}{3}]. \]

Case 2: \( y = 1 \)

If \( y = 1 \), the coefficient of \( x^2 \) in equation (1) vanishes, and the equation reduces to:

\[ x + 1 = 0. \]

This implies:

\[ x = -1. \]

For \( x = -1 \), substituting into \( f(x) \):

\[ f(-1) = \frac{(-1)^2}{(-1)^2 + (-1) + 1} = \frac{1}{1} = 1. \]

Thus, \( y = 1 \) is a valid output when \( x = -1 \).

Combining the two cases, the range of \( f(x) = \frac{x^2}{x^2 + x + 1} \) is:

\[ \left[0, \frac{4}{3}\right]. \]

Using Calculus to Find the Range of Functions

The range of a function can also be determined using calculus, specifically by finding the critical points through differentiation and analyzing the maxima and minima of the function. We will explore this method in detail later.