Problem 41 - Calculus Single Variable

Question

The derivative \(f^{\prime}\) of a function \(f\) is given. Determine and classify all local extrema of \(f\). $$ f^{\prime}(x)=\left(x^{2}-4\right)^{5} $$

Step-by-Step Solution

Answer

There are no local extrema for the function because the derivative does not change sign at \(x = -2\) or \(x = 2\).

Step 1: Determine Critical Points

Find the critical points by setting the derivative equal to zero and solving for \(x\). Since \(f^{\prime}(x) = (x^{2} - 4)^{5}\), set \((x^{2} - 4)^{5} = 0\). This implies \(x^{2} - 4 = 0\), which leads to \(x^{2} = 4\). Solving this gives the critical points \(x = 2\) and \(x = -2\).

Step 2: Analyze the Sign of the Derivative

To classify the critical points, analyze the sign of \(f^{\prime}(x)\) around each critical point. The derivative \(f^{\prime}(x) = (x^{2} - 4)^{5}\) is positive when \(x\) is less than \(-2\), positive when \(-2 < x < 2\), and positive when \(x > 2\). Thus, the derivative does not change sign across these intervals.

Step 3: Classify Local Extrema

Since the sign of \(f^{\prime}(x)\) does not change as \(x\) crosses critical points \(x = 2\) and \(x = -2\), there is no local maximum or minimum at these points. When the derivative remains positive, it indicates that the function is strictly increasing across these critical points.

Key Concepts

Understanding Local Extrema

Local extrema refer to the points on a graph of a function where the function reaches a local maximum or a local minimum.
A local maximum is a point where the function value is higher than nearby points, while a local minimum is where it is lower.
Identifying these points is crucial as it helps understand the general behavior or shape of the graph.To find local extrema, we usually rely on the function's derivative, denoted as \( f'(x) \).
This derivative tells us how the function is changing at different points, hinting at where the extrema may be.
Critical points, where local extrema can occur, are typically where the derivative is zero or undefined.In this exercise, solving \( (x^2 - 4)^5 = 0 \) gives critical points \( x = 2 \) and \( x = -2 \).
However, determining whether these are local maxima or minima requires further analysis.

Exploring Derivative Analysis

Derivative analysis involves understanding how the derivative of a function behaves in relation to the function itself.
It's vital for classifying critical points into local maxima, minima, or points of inflection.
The derivative at a certain point gives the slope of the tangent line to the function at that point.For instance, if \( f'(x) > 0 \), it implies the function is increasing at that section.
If \( f'(x) < 0 \), the function is decreasing.
This simple analysis helps us understand whether the function is rising or falling as we move along the x-axis.In the given example, the derivative \( (x^2 - 4)^5 \) does not change sign around the critical points.This indicates the function consistently increases on either side of the critical points, rather than having clear peaks (maxima) or troughs (minima).

The Impact of the Sign of the Derivative

Understanding the sign of a derivative is fundamental in finding local extrema.
It tells us about the increasing or decreasing nature of the function over intervals.
Here's a simple rule:

• If the derivative changes from positive to negative at a critical point, the function has a local maximum there.
• If it changes from negative to positive, there's a local minimum.
• If there's no sign change, there is neither a local maximum nor minimum at that point.

In our case, the derivative \( (x^2 - 4)^5 \) was always positive in intervals around the critical points \( x = 2 \) and \( x = -2 \), indicating a consistently increasing function.
This lack of sign change means there are no local extrema at these points, as confirmed by not observing any switches in direction (increasing to decreasing or vice versa) on the graph near these critical values.