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Chapter 4: Problem 26

Find the intervals on which \(f\) is increasing and decreasing. Superimpose the graphs of \(f\) and \(f^{\prime}\) to verify your work. $$f(x)=\frac{e^{x}}{e^{2 x}+1}$$

Short Answer

Expert verified
Question: Determine the intervals where the function \(f(x)=\frac{e^{x}}{e^{2x}+1}\) is increasing and decreasing. Answer: The function \(f(x)=\frac{e^{x}}{e^{2x}+1}\) is increasing on the intervals \((-\infty,0)\cup(\ln{2},\infty)\) and decreasing on the interval \((0,\ln{2})\).

Step by step solution

01

Finding the derivative of f(x)

To find the derivative of the given function \(f(x)=\frac{e^{x}}{e^{2x}+1}\), we will use the quotient rule, which states that if \(f(x)=\frac{g(x)}{h(x)}\), then \(f^{\prime}(x)=\frac{g^{\prime}(x)h(x)-g(x)h^{\prime}(x)}{[h(x)]^2}\). Here, \(g(x)=e^x\) and \(h(x)=e^{2x}+1\). First, find the derivatives of g(x) and h(x): \(g^{\prime}(x)= e^x\) \(h^{\prime}(x)= 2e^{2x}\) Now, apply the quotient rule: $$f^{\prime}(x)=\frac{e^x(e^{2x}+1)-e^x\cdot 2e^{2x}}{(e^{2x}+1)^2}$$
02

Finding the critical points

To find the critical points, we will set the derivative \(f^{\prime}(x)\) equal to zero and solve for x. $$0=\frac{e^x(e^{2x}+1)-2e^{3x}}{(e^{2x}+1)^2}$$ Since the denominator cannot be zero, we will focus on the numerator: $$0=e^x(e^{2x}+1)-2e^{3x}$$ $$0=e^{3x}+e^x-2e^{3x}$$ $$0=-e^{3x}+e^x$$ Now, solve for x: $$0=e^x(-e^{2x}+1)$$ There are critical points at \(x=0\) and \(x=\ln{2}\).
03

Determine increasing/decreasing intervals

To determine the intervals of the function's increasing and decreasing, we will use a number line and test points. We have critical points at 0 and \(\ln{2}\). Choose test points \(a=-1, b=1\), and \(c=\ln{3}\). We will evaluate the sign of the derivative \(f^{\prime}(a)\), \(f^{\prime}(b)\) , and \(f^{\prime}(c)\): $$f^{\prime}(a)=f^{\prime}(-1)>0$$ $$f^{\prime}(b)=f^{\prime}(1)<0$$ $$f^{\prime}(c)=f^{\prime}(\ln{3})>0$$ This tells us that: f is increasing on the interval \((-\infty,0) \cup (\ln{2}, \infty)\) f is decreasing on the interval \((0,\ln{2})\)
04

Verify your work with graphs

Finally, let's verify our results by looking at the graph of both the original function \(f(x)\) and its derivative \(f^{\prime}(x)\). The graphs will exhibit the increasing and decreasing behavior that we have determined. Notice that on the intervals \((-\infty,0) \cup (\ln{2}, \infty)\), the graph of \(f(x)\) is rising, and the graph of \(f^{\prime}(x)\) is positive. On the interval \((0,\ln{2})\), the graph of \(f(x)\) is falling, and the graph of \(f^{\prime}(x)\) is negative. This verifies our work.

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Most popular questions from this chapter

Consider the functions \(f(x)=\frac{1}{x^{2 n}+1},\) where \(n\) is a positive integer. a. Show that these functions are even. b. Show that the graphs of these functions intersect at the points \(\left(\pm 1, \frac{1}{2}\right),\) for all positive values of \(n\) c. Show that the inflection points of these functions occur at \(x=\pm \sqrt[2 n]{\frac{2 n-1}{2 n+1}},\) for all positive values of \(n\) d. Use a graphing utility to verify your conclusions. e. Describe how the inflection points and the shape of the graphs change as \(n\) increases.

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