Problem 37 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $$ f(x)=\left\\{\begin{array}{ll}{4-x^{2},} & {x \leq 0} \\ {-2 x,} & {x>0}\end{array}\right. $$

Step-by-Step Solution

Answer

The critical number of the function \(f(x)\) is \(x=0\). The function is decreasing on the interval \((-\infty,0]\), hence there is a relative maximum at \(x=0\).

Step 1: Find the critical numbers of \(f\)

The critical numbers of a function are the values of \(x\) where its derivative is either zero or undefined. We start by finding the derivative of \(f(x)\), separately for \(x \leq 0\) and \(x > 0\). For \(x \leq 0\), \(f'(x) = -2x\), and this is undefined for \(x > 0\). Thus, the critical numbers are the union of the solutions of \(f'(x) = 0\) and where \(f'(x)\) is undefined, which gives us the critical numbers as \(x=0\).

Step 2: Determine the intervals of increase or decrease.

For the function to be increasing, the derivative must be positive, and for it to be decreasing, the derivative must be negative. For \(x \leq 0\), \(f'(x) = -2x \leq 0\), the function is decreasing. For \(x > 0\), since \(f'(x)\) is undefined. Hence, overall, the function is decreasing on the interval \((-\infty, 0]\).

Step 3: Apply the First Derivative Test

We have one critical number, \(x = 0\), and one interval to consider, \((-\infty, 0]\). We see that the function is decreasing before it reaches \(x = 0\) and undefined afterwards. This means that \(x = 0\) is a relative maximum.

Step 4: Confirm the Solution with a Graph

Plotting this function with a graphing utility will show a declining line to the left of \(x = 0\) (including x=0) and a declining line to the right of \(x = 0\), confirming our results.

Key Concepts

Critical Numbers

Critical numbers are fundamental in calculus as they help identify points where important changes occur in a function's behavior. A critical number for a function can be an input where the function's first derivative is zero or undefined. These points might indicate potential maxima, minima, or points of inflection.

For the function given in the exercise, \[ f(x) = \begin{cases} 4-x^{2}, & x \leq 0 \ -2x, & x > 0 \end{cases}\]we determine the critical numbers by first calculating the derivative for each piece of the function. The derivative for the first piece, when \(x \leq 0\), is \(f'(x) = -2x\). For the second piece, when \(x > 0\), \(f'(x)\) is constant as \(-2\), but does not change, so it is undefined at \(x = 0\). Thus, the critical number in this case is \(x = 0\).

Finding critical numbers is a critical step in determining the function's behavior on a graph, signaling where it might change direction.

Increasing and Decreasing Intervals

The intervals on which a function is increasing or decreasing provide essential insights into its behavior. The derivative of a function helps us identify these intervals by showing where the function slopes upwards or downwards.

For \[ f(x) = \begin{cases} 4-x^{2}, & x \leq 0 \ -2x, & x > 0 \end{cases}\]the derivative \(f'(x) = -2x\) gives us a clue about where the function is decreasing when \(x \leq 0\) because the derivative is negative. Thus, the function is decreasing on \((- fty, 0]\).

For \(x > 0\), though the derivative doesn't clearly apply, since it's undefined, the lack of a positive derivative indicates the absence of increasing behavior there as well. Therefore, the function is never increasing.

Relative Extrema

Relative extrema refer to points in the function where the function value is a minimum or maximum relative to nearby points. Using the First Derivative Test, we can determine these extrema by analyzing the critical numbers and the function's behavior around these points.

Given a critical number \(x = 0\) for our piecewise function:\[ f(x) = \begin{cases} 4-x^{2}, & x \leq 0 \ -2x, & x > 0 \end{cases}\]the test shows that the function decreases as it approaches \(x = 0\) from the left. Since \(f'(x)\) is not defined after \(x = 0\) in terms of having a change from negative to positive or vice versa, \(x = 0\) indicates a relative maximum. Thus, we conclude that there's a relative maximum at \(x = 0\).

Understanding relative extrema aids in forming a complete picture of the function's graph.

Piecewise Function

A piecewise function is a function defined by different expressions depending on the domain segment. It can simplify modeling scenarios where the behavior changes at certain points, like the one we are analyzing here.

The function\[ f(x) = \begin{cases} 4-x^{2}, & x \leq 0 \ -2x, & x > 0 \end{cases}\]exemplifies a piecewise function with different expressions over the partitioned domain. This piecewise format provides flexibility in handling various equations as needed for different subdomains, simplifying complex real-world modeling.

Understanding piecewise functions equips us to interpret situations with multiple criteria effectively and allows us to isolate and analyze separate behaviors within the same overarching function.