Ytukyg

I don't know how to approach this problem:

For what values of the numbers $a$ and $b$ does the function $f(x)=axe^left(bx^2right)$ have the maximum value $f(2)=8?$

Thanks so much for any help!

Hint:

1) Equate derivative at $x=2$ to zero. You will find $b=-frac18$.

2) Use the given condition $f(2)=8$ to find $a=4e^{1/2}$.

3) Make sure the second derivative is less than zero. You will get $a>0$.

The extreme value of the function happens at $x_0$ such that $f'(x_0)=0$. Then, you can simply take derivative to get $$f'(x)=ae^{bx^2}+2abx^2e^{bx^2}=0$$
You can solve for $x$:

begin{align*}
ae^{bx^{2}}+2abx^{2}e^{bx^{2}}&=0\2abx^{2}e^{bx^{2}}&=-ae^{bx^{2}}\2bx^{2}&=-1\x^{2}&=-frac{1}{2b}\&=pmsqrt{-frac{1}{2b}}
end{align*}

This means that $x=sqrt{-frac{1}{2b}}$ and $x=-sqrt{-frac{1}{2b}}$ are two points where the function attains extreme value.

Note that $sqrt{-frac{1}{2b}}$ is always positive, so if the maximum happens at $f(2)$, $sqrt{-frac{1}{2b}}=2$ and with some algebra, we can deduce that $b=-frac{1}{8}$

Now we have that $f(2)=2ae^{-frac{1}{8}2^2}=8$. Then, we can simply find begin{align*}2ae^{-frac{1}{8}2^{2}}&=8\ae^{-frac{1}{2}}&=4\a&=4e^{1/2}\&=4sqrt{e}end{align*}