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Chapter 4: Problem 37

Find each critical point \(c\) of the given function \(f\). Then use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=(x-1)^{1 / 3}(x+5)^{2 / 3} $$

Short Answer

Expert verified
Critical points are at \(x = 1\) and \(x = -5\). \(f(-5)\) is a local max; \(f(1)\) is a local min.

Step by step solution

01

Find the Derivative of the Function

To locate the critical points, we need the derivative of the function. Using the product rule and the chain rule, the derivative of \(f(x)=(x-1)^{1/3}(x+5)^{2/3}\) is calculated. Let \(u(x) = (x-1)^{1/3}\) and \(v(x) = (x+5)^{2/3}\), then:\[f'(x) = u'(x)v(x) + u(x)v'(x)\]\[u'(x) = \frac{1}{3}(x-1)^{-2/3}\], \[v'(x) = \frac{2}{3}(x+5)^{-1/3}\].Hence, \[f'(x) = \frac{1}{3}(x-1)^{-2/3}(x+5)^{2/3} + \frac{2}{3}(x-1)^{1/3}(x+5)^{-1/3}\].
02

Find Critical Points

Critical points occur where \( f'(x) = 0 \) or where the derivative does not exist. Begin by setting the derivative equal to zero \( f'(x)=0 \) and finding points where the derivative is undefined.Factor and solve:\[0 = \frac{1}{3}(x-1)^{-2/3}(x+5)^{2/3} + \frac{2}{3}(x-1)^{1/3}(x+5)^{-1/3}\].After simplification, observe critical points at \( x = 1 \) and \( x = -5 \) when the derivative is undefined.
03

Evaluate the First Derivative Test

To use the First Derivative Test, test the intervals created by the critical points \(x = 1\) and \(x = -5\). Choose test points in each interval to determine the sign of \(f'(x)\).1. Interval \((-\infty,-5)\): Choose \(x = -6\), evaluate \(f'(-6) > 0\).2. Interval \((-5,1)\): Choose \(x = 0\), evaluate \(f'(0) < 0\).3. Interval \((1,\infty)\): Choose \(x = 2\), evaluate \(f'(2) > 0\).
04

Determine the Nature of each Critical Point

Using the First Derivative Test:- At \(x = -5\), \( f'(x) \) changes from positive to negative, making \(f(-5)\) a local maximum.- At \(x = 1\), \( f'(x) \) changes from negative to positive, making \(f(1)\) a local minimum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative Test
The First Derivative Test is a handy tool used to determine whether a particular point on a function is a local maximum, minimum, or neither. Once you have a critical point, which happens when the derivative is zero or doesn't exist, you can use this test to understand the behavior of the function around that point.
  • First, divide the number line into intervals using the critical points.
  • Second, select test points from each interval and evaluate the derivative at these points.
  • If the derivative changes from positive to negative, it's a local maximum at that point.
  • If the derivative changes from negative to positive, it's a local minimum at that point.
In our problem, for example, at the critical point \( x = -5 \), the derivative \( f'(x) \) changes from positive to negative, indicating a local maximum. Whereas at \( x = 1 \), it changes from negative to positive, indicating a local minimum.
Derivative Calculation
Derivative calculation is a fundamental skill in calculus used to measure how a function changes. Finding the derivative of a function gives you a new function, which tells you the rate of change or the slope of the original function at any point. To find critical points, you first need to find the derivative.
In general, you find derivatives using various rules such as the product rule, quotient rule, and chain rule. For our function \( f(x) = (x-1)^{1/3}(x+5)^{2/3} \), we begin by calculating the derivative using the product and chain rule.
  • The purpose is to pinpoint where the function's slope is zero, which occurs at its peaks or troughs, or where it does not exist.
  • These are often the spots where critical points occur, guiding us to the local peaks and valleys of the function.
Product Rule
The product rule is applied when you need to take the derivative of a product of two functions. If you have two functions, \( u(x) \) and \( v(x) \), the derivative of their product \( f(x) = u(x) \times v(x) \) is found using the product rule.
The formula is: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] This allows us to "distribute" the derivative through the product, ensuring we consider changes in both \( u(x) \) and \( v(x) \).
In the domain of our original problem:
  • We let \( u(x) = (x-1)^{1/3} \) and \( v(x) = (x+5)^{2/3} \),
  • Then applied the product rule to find \( f'(x) \) by taking derivatives of \( u(x) \) and \( v(x) \), as shown in the step-by-step solution.
Thus, breaking down a product into simpler parts allows us to handle complex derivatives smoothly.
Chain Rule
The chain rule is essential when dealing with composite functions, where one function is inside another. It allows you to find the derivative of a function by focusing on the outer function first and then the inner function.
If you have a composite function \( f(g(x)) \), its derivative is \[ f'(g(x)) \cdot g'(x) \]To successfully apply the chain rule, you need to recognize the "layers" of the function. In our exercise:
  • Each part \( (x-1)^{1/3} \) and \( (x+5)^{2/3} \) required the chain rule for its derivative.
  • We first differentiated the outer function, keeping the inside unchanged. Then multiplied by the derivative of the inside function.
This step ensures we holistically assess changes in more complicated nested functions, allowing us to precisely determine their rate of change.

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Most popular questions from this chapter

For each given function \(f,\) graph the function \(S(x)=\) signum \(\left(f^{\prime}(x)\right)\) for \(x\) in the given interval \(I .\) Use the graph of \(S\) to determine and classify the local extrema of \(f\). a. \(f(x)=x^{5}-12 x^{4}+55 x^{3}-120 x^{2}+124 x-48\) \(I=[0.8,4.1]\) b. \(f(x)=4 x^{4}-24 x^{3}+51 x^{2}-44 x+12, \quad I=[0.3,2.6]\) c. \(f(x)=x^{4}-x^{3}-7 x^{2}+x+6, \quad I=[-2.4,3.2]\)

Without using calculus, prove that the composition of two increasing functions is increasing. Now assume that \(f\) and \(g\) are differentiable functions with positive derivatives and that \(g \circ f\) is defined. Use calculus to show that \(g \circ f\) is increasing.

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Let \(f: \mathbb{R} \rightarrow(0, \infty)\) be a differentiable function. Compare the local extrema of \(f\) with those for \(1 / f\). The local maxima for \(f\) become what for \(1 / f ?\) The local minima for \(f\) become what for 1/f?
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