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Chapter 1: Problem 101

Find the interval(s) for which \(f^{\prime}(x)\) is positive. Use the derivative to help explain why \(k(x)=\frac{1}{x^{2}}\) decreases for all \(x\) in \((0, \infty)\).

Short Answer

Expert verified
\(k^{\prime}(x)\) is never positive; \(k(x)\) is decreasing for \(x \in (0, \infty)\).

Step by step solution

01

Find the derivative of k(x)

Given the function \(k(x) = \frac{1}{x^2}\), we need to first find the derivative \(k^{\prime}(x)\) using the power rule. Rewriting \(k(x)\) gives \(k(x) = x^{-2}\). Differentiating, we have: \[ k^{\prime}(x) = -2x^{-3} = -\frac{2}{x^3} \]
02

Determine where the derivative is positive

To find where \(k^{\prime}(x)\) is positive, we solve the inequality:\[-\frac{2}{x^3} > 0\]This inequality implies that there is no positive solution since the negative fraction can never be greater than zero. This confirms that \(k^{\prime}(x)\) is never positive.
03

Explain why k(x) decreases for x > 0

The derivative \(k^{\prime}(x) = -\frac{2}{x^3}\) is always negative for \(x > 0\). A negative derivative indicates that the function \(k(x)\) is decreasing on the interval \((0, \infty)\). This means for any \(x > 0\), as \(x\) increases, \(k(x)\) decreases.

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Key Concepts

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

Understanding Derivatives
In calculus, a derivative is a way to show the rate at which a function is changing at any given point. Think of it as a way to understand the "instantaneous speed" of a function. If you've ever driven a car and looked at the speedometer to see how fast you're going at an exact moment, you're using a concept similar to derivatives.
Just like with speed, if the derivative value is positive, the function is increasing at that point. If the derivative is negative, the function is decreasing. In the exercise we're looking at, the function is always decreasing because its derivative is negative for all concerned values of the variable.
Applying the Power Rule
The power rule is a handy tool for finding the derivative of functions that are expressed as powers of a variable. It's quick and straightforward. To use the power rule, multiply the variable's exponent by the coefficient and reduce the exponent by one.
For example, if you have a function such as \( x^n \), the derivative is \( nx^{n-1} \).
In this exercise, the function \( k(x) = \frac{1}{x^2} \) is rewritten as \( x^{-2} \). Applying the power rule gives a derivative of \( -2x^{-3} \).
This result helps us understand how this function behaves, specifically that its negative sign implies a downward slope.
Identifying Decreasing Functions
A decreasing function, as the name suggests, is a function that goes downwards as you move along the x-axis from left to right. It means the output, or \( y \)-value, of a function gets smaller as the input \( x \) gets larger.
The derivative tells us a lot about whether a function is increasing or decreasing. Specifically, if a function's derivative is negative across its domain, the function itself is decreasing.
  • The derivative of \( k(x) = \frac{1}{x^2} \) is \( -\frac{2}{x^3} \).
  • As it is negative for all \( x > 0 \), \( k(x) \) decreases in the interval \((0, \infty)\).
This negative derivative \( -\frac{2}{x^3} \) confirms that \( k(x) \) behaves like a decreasing function for all \( x > 0 \). Each increase in \( x \) leads to a smaller value of \( k(x) \).

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Most popular questions from this chapter

Below are the steps in the simplification of the difference quotient for \(f(x)=\sqrt{x}\) (see Example 8). Provide a brief justification for each step. \(\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\) a) \(=\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot\left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\) b) \(=\frac{x+h+\sqrt{x} \sqrt{x+h}-\sqrt{x} \sqrt{x+h}-x}{h(\sqrt{x+h}+\sqrt{x})}\) c) \(=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\) d) \(=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\) e) \(=\frac{1}{\sqrt{x+h}+\sqrt{x}}\)

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 2}\left(\frac{x^{3}+5 x-18}{2 x^{2}-8}\right) $$

It has been shown that the home range, in hectares, of a carnivorous mammal weighing \(w\) grams can be approximated by $$ H(w)=0.11 w^{1.36} $$ (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. \(216,\) Reading, MA: Addison-Wesley, 1973 ; and Harestad, A. S., and Bunnel, F. L., "Home Range and Body Weight-A Reevaluation," Ecology, Vol. 60, No. 2, pp. 405-418.) a) Find the average rate at which a carnivorous mammal's home range increases as the animal's weight grows from \(500 \mathrm{~g}\) to \(700 \mathrm{~g}\). b) Find \(\frac{H(300)-H(200)}{300-200}\). What does this rate represent?

The equation $$ S(r)=\frac{1}{r^{4}} $$ can be used to determine the resistance to blood flow, \(S\). of a blood vessel that has radius \(r\), in millimeters (mm). a) Find the rate of change of resistance with respect to \(r\), the radius of the blood vessel. b) Find the resistance at \(r=1.2 \mathrm{~mm}\). c) Find the rate of change of \(S\) with respect to \(r\) when \(r=0.8 \mathrm{~mm}\)

The cost of sending a large envelope via U.S. first-class mail in 2014 was \(\$ 0.98\) for the first ounce and \(\$ 0.21\) for each additional ounce (or fraction thereof). (Source: www.usps.com.) If \(x\) represents the weight of a large envelope, in ounces, then \(p(x)\) is the cost of mailing it, where $$ \begin{array}{l} p(x)=\$ 0.98, \text { if } \quad 0 < x \leq 1, \\ p(x)=\$ 1.19, \text { if } \quad 1 < x \leq 2, \\ p(x)=\$ 1.40, \text { if } 2 < x \leq 3, \end{array} $$ and so on, up through 13 ounces. The graph of \(p\) is shown below. Using the graph of the postage function, find each of the following limits, if it exists. $$ \lim _{x \rightarrow 3} p(x) $$
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