Problem 10 - Calculus: Early Transcendentals

Question

Find all critical points and identify them as local maximum points, local minimum points, or neither. $$ f(x)=\left|x^{2}-121\right| $$

Step-by-Step Solution

Answer

Critical points: \( x = 0 \) (local max), \( x = \pm 11 \) (transition points).

Step 1: Understand the Function

The function given is \( f(x) = |x^2 - 121| \). This function is an absolute value function, which can be challenging due to the piecewise nature that arises from the absolute value.

Step 2: Define the Piecewise Function

Since \( |x^2 - 121| \) is based on the absolute value, it can be divided into two cases: \( f(x) = x^2 - 121 \) when \( x^2 - 121 \geq 0 \) and \( f(x) = -(x^2 - 121) \) when \( x^2 - 121 < 0 \). This simplifies to:- \( f(x) = x^2 - 121 \) for \( x \geq 11 \) or \( x \leq -11 \) - \( f(x) = 121 - x^2 \) for \( -11 < x < 11 \).

Step 3: Find the Derivative

To find critical points, calculate \( f'(x) \) by finding the derivative of each piece. For \( x^2 - 121 \), the derivative is \( f'(x) = 2x \).For \( 121 - x^2 \), the derivative is \( f'(x) = -2x \).

Step 4: Find Critical Points

Critical points occur where the derivative is zero or undefined. Solve:- \( 2x = 0 \) gives \( x = 0 \).- \( -2x = 0 \) also gives \( x = 0 \).Check the domain of each piece to confirm \( x = 0 \) as a critical point (it lies within both domains). Also, evaluate the endpoints \( x = 11 \) and \( x = -11 \) for possible critical points due to changes in piecewise cases.

Step 5: Identify the Nature of the Critical Points

Evaluate the second derivative for each piece or use the first derivative test:- For \( x < -11 \) and \( x > 11 \), \( f(x) = x^2 - 121 \). The second derivative \( f''(x) = 2 > 0 \) indicates local minima.- For \( -11 < x < 11 \), \( f(x) = 121 - x^2 \). The second derivative \( f''(x) = -2 < 0 \) indicates local maxima.Since \( x = 0 \) lies in the interval \(-11 < x < 11\), it is a local maximum. At \( x = -11 \) and \( x = 11 \), since the second derivative switches signs but remains positive nearby, they can be considered points where the function changes but must be evaluated for overall function behavior.

Step 6: Conclude Critical Points

The critical points are \( x = 0 \), \( x = 11 \), and \( x = -11 \). At \( x = 0 \), there is a local maximum. At \( x = 11 \) and \( x = -11 \), these points act as locations where the behavior of the absolute value function changes but do not conform to typical maxima/minima definitions without checking continuity and context over the domain.

Key Concepts

Derivative

Derivatives are a fundamental concept in calculus, used to determine how a function changes at any given point. They provide the rate of change of a function with respect to its variable.

The derivative of a function gives us the slope of the tangent line at any point on the graph. In the case of our exercise function, the derivative is calculated separately for each piece of the piecewise function.

• For the piece where the function is expressed as \( x^2 - 121 \), the derivative is \( f'(x) = 2x \).
• For the piece where the function is \( 121 - x^2 \), the derivative is \( f'(x) = -2x \).

This distinction is necessary because the function is approached differently on intervals around its critical points. Understanding derivatives helps us find critical points by setting the derivatives equal to zero or identifying where they do not exist. These critical points are where the function could have local maxima or minima.

Piecewise Function

A piecewise function is a function composed of different sub-functions, each applied to a certain part of the function's domain. The key challenge with piecewise functions is managing the different expressions and conditions defining each piece without a break.

For our exercise, the absolute value of \( x^2 - 121 \) leads to a piecewise definition:

• When \( x^2 - 121 \geq 0 \), it simplifies to \( f(x) = x^2 - 121 \).
• When \( x^2 - 121 < 0 \), it switches to \( f(x) = 121 - x^2 \).

This essentially forms a boundary at \( x = 11 \) and \( x = -11 \) where the nature of the function changes, due to the absolute value opening a shift in how the value of \( f(x) \) is calculated. Analyzing these separate domains individually is crucial to understanding the overall behavior.

Absolute Value Function

Absolute value functions are often encountered in mathematical analysis due to their piecewise nature. The absolute value of a number represents its distance from zero on the number line, regardless of its direction.

When dealing with an absolute value function like \( f(x) = |x^2 - 121| \), it inherently constructs a piecewise function:

• It retains the original expression \( x^2 - 121 \) whenever the squared term is greater or equal to 121.
• If the squared term is less than 121, the result flips to negative, expressing \( 121 - x^2 \).

The challenge is that these functions must be evaluated separately in their different intervals. Absolute values give rise to critical points and require careful handling to understand their effects on function behavior.

Second Derivative Test

The Second Derivative Test is a powerful tool in calculus to determine if a function's graph has a local maximum, a local minimum, or a point of inflection at a critical point.

In this method, you evaluate the second derivative at the critical points determined by setting the first derivative to zero. Here's how it works in the context of our exercise:

• For \( f(x) = x^2 - 121 \) where \( x < -11 \) or \( x > 11 \), the second derivative is \( f''(x) = 2 \), which is positive. Therefore, these regions indicate local minima.
• For \( f(x) = 121 - x^2 \) when \( -11 < x < 11 \), the second derivative is \( f''(x) = -2 \), a negative value, indicating local maxima.

The Second Derivative Test simplifies identifying the nature of critical points by determining the concavity of the function at those points. It makes assessing whether a point is a maximum or minimum much easier and less laborious than graphing.