## Calculus: Early Transcendentals 8th Edition

$$a=\frac{\sqrt e}{2}$$ and $$b=-\frac{1}{8}$$
$$f(x)=axe^{bx^2}$$ 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.