Absolute Maximum and Minimum of a Function
Step-by-Step Calculus Examples

Learn how to find the absolute maximum and absolute minimum of a function using first derivatives, critical points, and interval evaluation. This guide includes graphical interpretations to help visualize the concepts.

Understanding Absolute Extrema

Definition: Values of \(x\) in the domain of a function \(f\) where \(f'(x) = 0\) or \(f'(x)\) is undefined are called critical points of \(f\).

Theorem: For a continuous function on a closed interval \([a, b]\), there exist values \(x_1\) and \(x_2\) in \([a, b]\) such that: \[ f(x_1) = \min f(x), \quad f(x_2) = \max f(x), \quad \text{for all } x \in [a, b]. \]

Let’s explore examples that demonstrate how absolute extrema can occur at critical points or interval endpoints.

Example 1: Extrema at Stationary Points

When \(f'(x_0) = 0\) and the sign of \(f'(x)\) changes, \(x_0\) may correspond to a local (and possibly absolute) maximum or minimum.

Example 2: Extrema Where Derivative is Undefined

Extrema may occur where the derivative does not exist.

Example 3: Extrema at Endpoints

Consider \(f(x) = x^3 - 2x^2\) on \([-1, 5/2]\). Local extrema exist, but the absolute extrema occur at the endpoints:

General Steps to Find Absolute Extrema

1. Find the first derivative \(f'(x)\).
2. Determine the critical points (where \(f'(x) = 0\) or undefined).
3. Evaluate \(f(x)\) at all critical points and endpoints of the interval.
4. The largest value is the absolute maximum, and the smallest is the absolute minimum.

Example 4: Absolute Extrema of a Quadratic Function

Solution to Example 4

Step 1: Compute the First Derivative
To find the absolute maximum and minimum of the function, we begin by calculating the first derivative of \( f(x) = -x^2 + 2x - 2 \):
\[ f'(x) = -2x + 2 \]

Step 2: Determine the Critical Points
Set the derivative equal to zero to find the critical points:
\[ -2x + 2 = 0 \quad \Rightarrow \quad x = 1 \]
Since the first derivative is defined for all \( x \) in the interval \([-2, 3]\), and \( x = 1 \) lies within that interval, it is a valid critical point.

Step 3: Evaluate the Function at Endpoints and Critical Points
Now evaluate the original function at the endpoints and at the critical point \( x = 1 \):
\[ f(-2) = -(-2)^2 + 2(-2) - 2 = -4 - 4 - 2 = -10 \]
\[ f(1) = -(1)^2 + 2(1) - 2 = -1 + 2 - 2 = -1 \]
\[ f(3) = -(3)^2 + 2(3) - 2 = -9 + 6 - 2 = -5 \]

Conclusion:
- The absolute maximum value is \( f(1) = -1 \), which occurs at \( x = 1 \).
- The absolute minimum value is \( f(-2) = -10 \), which occurs at \( x = -2 \).

The graph below illustrates the function, showing the critical point and endpoints, and highlighting the absolute minimum and maximum values:

Example 5: Finding Absolute Extrema of a Polynomial Function on a Closed Interval

Determine the absolute maximum and minimum values of the function \( f(x) = \dfrac{1}{4} x^4 + \dfrac{1}{3} x^3 - x^2 \) on the closed interval \( [-1, 1] \).

Solution to Example 5

Step 1: Compute the First Derivative

Differentiate the function to find its critical points:

\( f'(x) = x^3 + x^2 - 2x \)

Step 2: Find Critical Points by Solving \( f'(x) = 0 \)

Set the derivative equal to zero:

\( x^3 + x^2 - 2x = 0 \Rightarrow x(x - 1)(x + 2) = 0 \)

The solutions are \( x = 0 \), \( x = 1 \), and \( x = -2 \). However, only \( x = 0 \) and \( x = 1 \) lie within the interval \( [-1, 1] \).

So the critical points in the domain are \( x = 0 \) and \( x = 1 \).

Step 3: Evaluate the Function at Endpoints and Critical Points

• \( f(-1) = \dfrac{1}{4}(-1)^4 + \dfrac{1}{3}(-1)^3 - (-1)^2 = \dfrac{1}{4} - \dfrac{1}{3} - 1 = -\dfrac{13}{12} \)
• \( f(0) = \dfrac{1}{4}(0)^4 + \dfrac{1}{3}(0)^3 - (0)^2 = 0 \)
• \( f(1) = \dfrac{1}{4}(1)^4 + \dfrac{1}{3}(1)^3 - (1)^2 = \dfrac{1}{4} + \dfrac{1}{3} - 1 = -\dfrac{5}{12} \)

Step 4: Identify Absolute Maximum and Minimum

• Absolute maximum value: \( f(0) = 0 \) at \( x = 0 \)
• Absolute minimum value: \( f(-1) = -\dfrac{13}{12} \) at \( x = -1 \)

The graph of the function \( f(x) \) is shown below, illustrating the critical points, endpoints, and the locations of the absolute maximum and minimum.

Example 6: Finding the Absolute Maximum and Minimum of a Function Involving Logarithms

Find the absolute maximum and minimum values of the function \( f(x) = x^2 \ln(x) - 1 \) on the closed interval \( [0.5 , 2] \).

Solution to Example 6

Step 1: Compute the First Derivative

We differentiate using the product rule: \[ f(x) = x^2 \ln(x) - 1 \] \[ f'(x) = \dfrac{d}{dx}(x^2 \ln(x)) = 2x \ln(x) + x \]

Step 2: Find the Critical Points

Set the first derivative equal to zero: \[ f'(x) = 2x \ln(x) + x = x(2 \ln(x) + 1) = 0 \] Solve:

• \( x = 0 \) (excluded since it's not in the domain)
• \( 2 \ln(x) + 1 = 0 \Rightarrow \ln(x) = -\dfrac{1}{2} \Rightarrow x = e^{-1/2} \approx 0.6065 \)

Only \( x = e^{-1/2} \approx 0.6065 \) lies within the interval \( [0.5 , 2] \), so it's the only critical point in the domain.

Step 3: Evaluate the Function at Endpoints and Critical Point

• \( f(0.5) = (0.5)^2 \ln(0.5) - 1 = 0.25 \cdot (-0.6931) - 1 \approx -0.1733 - 1 = -1.1733 \)
• \( f(2) = (2)^2 \ln(2) - 1 = 4 \cdot 0.6931 - 1 = 2.772 - 1 = 1.772 \)
• \( f(e^{-1/2}) = (e^{-1}) \cdot (-\dfrac{1}{2}) - 1 = -\dfrac{1}{2e} - 1 \approx -0.1839 - 1 = -1.1839 \)

Final Answer

• Absolute maximum value: \( \boxed{1.772} \) at \( x = 2 \)
• Absolute minimum value: \( \boxed{-1.1839} \) at \( x = e^{-1/2} \approx 0.6065 \)

Example 7: Absolute Maximum and Minimum of an Absolute Value Function

Find the absolute maximum and minimum values of the function \( f(x) = |x^2 - 2x - 3| - x \) on the closed interval \([-1.1 , 4]\).

Solution to Example 7

Step 1: Rewrite and Differentiate the Function

Use the identity \( |u| = \sqrt{u^2} \) to rewrite the function:

\( f(x) = \sqrt{(x^2 - 2x - 3)^2} - x \)

Now differentiate:

\( f'(x) = \dfrac{(x^2-2x-3)(2x-2)}{\sqrt{(x^2-2x-3)^2}} - 1 = \dfrac{(x^2-2x-3)(2x-2) - |x^2-2x-3|}{|x^2-2x-3|} \)

Step 2: Find Critical Points

The derivative is undefined when the denominator is zero:

Set \( x^2 - 2x - 3 = 0 \) ⇒ \( (x+1)(x-3) = 0 \) ⇒ \( x = -1 \) and \( x = 3 \)

To find where \( f'(x) = 0 \), solve:

\( (x^2-2x-3)(2x-2) - |x^2-2x-3| = 0 \)     (Equation 1)

Case 1: \( x^2 - 2x - 3 < 0 \)

Then \( |x^2 - 2x - 3| = -(x^2 - 2x - 3) \), and Equation 1 becomes:

\( (x^2-2x-3)(2x-2) + (x^2-2x-3) = (x^2-2x-3)(2x-1) = 0 \)

Solutions: \( x = -1 \), \( x = 3 \), \( x = \dfrac{1}{2} \)

Case 2: \( x^2 - 2x - 3 > 0 \)

Then \( |x^2 - 2x - 3| = x^2 - 2x - 3 \), and Equation 1 becomes:

\( (x^2-2x-3)(2x-3) = 0 \)

Solutions: \( x = -1 \), \( x = 3 \), \( x = \dfrac{3}{2} \)

Verify Critical Points

• \( x = -1 \), \( x = 3 \): Derivative undefined — critical points.
• \( x = \dfrac{1}{2} \): Plug into numerator ⇒ result is 0 ⇒ valid critical point.
• \( x = \dfrac{3}{2} \): Plug into numerator ⇒ not 0 ⇒ not a critical point.

Conclusion: The critical points of \( f(x) \) are: \( x = -1 \), \( x = \dfrac{1}{2} \), and \( x = 3 \).

Step 3: Evaluate \( f(x) \) at Endpoints and Critical Points

• \( f(-1.1) = |(-1.1)^2 - 2(-1.1) - 3| - (-1.1) = 1.51 \)
• \( f(4) = |4^2 - 2 \cdot 4 - 3| - 4 = 1 \)
• \( f\left(\dfrac{1}{2}\right) = \left| \dfrac{1}{4} - 1 - 3 \right| - \dfrac{1}{2} = \dfrac{13}{4} = 3.25 \)
• \( f(-1) = |1 + 2 - 3| + 1 = 1 \)
• \( f(3) = |9 - 6 - 3| - 3 = -3 \)

Final Answer:

• Absolute Maximum: \( f(x) = 3.25 \) at \( x = \dfrac{1}{2} \)
• Absolute Minimum: \( f(x) = -3 \) at \( x = 3 \)

Graph of the Function

The graph of \( f(x) = |x^2 - 2x - 3| - x \) below shows the critical points and the endpoints of the interval \([-1.1 , 4]\), along with the absolute maximum and minimum values.

Example 8 Absolute Extrema of Functions with Rational Power

Find the absolute maximum and minimum of function \( f \) defined by \[ f(x) = (x-2)^{2/5} \;\; \text{on} \;\; [-3 , 4] \].

Solution to Example 8

Step 1: Compute the first derivative of the function
Given the function:
\( f(x) = (x - 2)^{2/5} \)

The first derivative is:
\[ f'(x) = \dfrac{2}{5}(x - 2)^{-3/5} = \dfrac{2}{5(x - 2)^{3/5}} \]

Step 2: Identify critical points
The derivative \( f'(x) \) has no zeros because the numerator is constant.
However, the derivative is undefined when the denominator is zero, i.e., when \( x = 2 \).
Therefore, \( x = 2 \) is a critical point because the derivative is not defined at that point.

Step 3: Evaluate f(x) at the endpoints and the critical point
Compute values of the function at key points within the interval \([-3, 4]\):

\[ f(-3) = ((-3) - 2)^{2/5} = (-5)^{2/5} \approx 1.90 \]
\[ f(4) = (4 - 2)^{2/5} = 2^{2/5} \approx 1.32 \]
\[ f(2) = (2 - 2)^{2/5} = 0^{2/5} = 0 \]

Step 4: Determine absolute maximum and minimum
From the evaluations above:

• Absolute minimum: 0 at \( x = 2 \)
• Absolute maximum: approximately 1.90 at \( x = -3 \)

Graphical Representation:
The graph of \( f(x) = (x - 2)^{2/5} \) over the interval \([-3, 4]\) confirms the critical point at \( x = 2 \), as well as the absolute minimum and maximum values.

More References and Links