Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises: 74


$$a=\frac{\sqrt e}{2}$$ and $$b=-\frac{1}{8}$$

Work Step by Step

$$f(x)=axe^{bx^2}$$ 1) First, we need to find $f'(x)$ and $f''(x)$ $$f'(x)=a(xe^{bx^2})'$$ $$f'(x)=a[x'e^{bx^2} + x(e^{bx^2})']$$ $$f'(x)=a[e^{bx^2}+xe^{bx^2}(bx^2)']$$ $$f'(x)=a[e^{bx^2}+xe^{bx^2}2bx]$$ $$f'(x)=ae^{bx^2}(1+2bx^2)$$ $$f''(x)=a[(e^{bx^2})'(1+2bx^2)+e^{bx^2}(1+2bx^2)']$$ $$f''(x)=a[2bxe^{bx^2}(1+2bx^2)+e^{bx^2}(4bx)]$$ $$f''(x)=a(2bxe^{bx^2})(1+2bx^2+2)$$ $$f''(x)=2abxe^{bx^2}(2bx^2+3).$$ 2) Now, the function has a maximum value $f(2)=1$, which means $x=2$ is a critical number of $f(x)$. Therefore, $$f'(2)=0$$ $$ae^{b(2^2)}(1+2b\times2^2)=0$$ $$ae^{4b}(1+8b)=0\hspace{1cm}(1).$$ 3) The function has a maximum value $f(2)=1$, also means that the point $(2,1)$ lies on the graph of the function $f$. In other words, $$f(2)=1$$ $$2ae^{b(2^2)}=1$$ $$ae^{4b}=\frac{1}{2}\hspace{1cm}(2).$$ 4) Now we consider $(1)$ $$ae^{4b}(1+8b)=0$$ We know that $e^{4b}\gt0$ for all $b\in R$, so $e^{4b}\ne0$. That means $$a=0\hspace{.5cm}or\hspace{.5cm}1+8b=0$$ $$a=0\hspace{.5cm}or\hspace{.5cm}b=-\frac{1}{8}.$$ However, consider $(2)$. It reveals that $a\ne0$. Since if $a=0$, then $$0e^{4b}=0\ne\frac{1}{2}.$$ Therefore, the only choice is that $$b=-\frac{1}{8}.$$ So $(2)$ would become $$ae^{4\times(-1/8)}=\frac{1}{2}$$ $$ae^{-\frac{1}{2}}=\frac{1}{2}$$ $$\frac{a}{\sqrt e}=\frac{1}{2}$$ $$a=\frac{\sqrt e}{2}$$ 5) Now the final step is to plug just found $a$ and $b$ back into $f''(x)$ to see if there really is a maximum value $f(2)=1.$ $$f''(x)=2\frac{\sqrt e}{2}(-\frac{1}{8})xe^{(-1/8)x^2}[2(-\frac{1}{8})x^2+3]$$ $$f''(x)=-\frac{\sqrt e}{8}xe^{-x^2/8}[-\frac{x^2}{4}+3]$$ $$f''(x)\approx-0.206xe^{-x^2/8}[-\frac{x^2}{4}+3]$$ Now we try $f''(2)$ here $$f''(2)\approx-0.206\times2e^{-2^2/8}[-\frac{2^2}{4}+3]$$ $$f''(2)\approx-0.206\times2e^{-1/2}[-1+3]$$ $$f''(2)\approx-0.206\times\frac{2}{\sqrt e}\times2$$ $$f''(2)\approx-0.206\times1.213\times2$$ $$f''(2)\approx-0.5\lt0$$ Since $f'(2)=0$ and $f''(2)\lt0$, according to the Second Derivative Test, there is a maximum value at $x=2$, which is $f(2)=1$. The found values of $a$ and $b$ are therefore true.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.