# Question #a621d

Dec 3, 2017

I am assuming missing parentheses to get $f \left(x\right) = a x {e}^{b {x}^{2}}$. Which makes $b = - \frac{1}{18}$ and $a = \frac{7}{3} \sqrt{e}$

#### Explanation:

$f ' \left(x\right) = a {e}^{b {x}^{2}} \left(1 + 2 b {x}^{2}\right) = 0$ at ${x}^{2} = - \frac{1}{2 b}$

So we need $x = \sqrt{- \frac{1}{2 b}} = 3$ and so,

$b = - \frac{1}{18}$

Now we have $f \left(x\right) = a x {e}^{- {x}^{2} / 18}$.

We also want $f \left(x\right) = 7$, so solve

$a \left(3\right) \left({e}^{- \frac{1}{2}}\right) = 7$ for $a = \frac{7}{3} {e}^{\frac{1}{2}}$