this is similar to section 3.7 problem 38: for the function y = 9xe^{4x}, determine the absolute maximum and…

Explanation:

Step1: Find the derivative using product - rule

The product - rule states that if $y = uv$, where $u = 9x$ and $v=e^{4x}$, then $y^\prime=u^\prime v + uv^\prime$. $u^\prime = 9$ and $v^\prime = 4e^{4x}$. So $y^\prime=9e^{4x}+9x\times4e^{4x}=9e^{4x}(1 + 4x)$.

Step2: Find the critical points

Set $y^\prime = 0$. Since $e^{4x}>0$ for all real $x$, we solve $1 + 4x=0$. $1+4x = 0$ gives $x=-\frac{1}{4}$.

Step3: Evaluate the function at the critical point and the limit as $x\to-\infty$

Evaluate $y$ at $x =-\frac{1}{4}$: $y=9\times(-\frac{1}{4})e^{4\times(-\frac{1}{4})}=-\frac{9}{4e}\approx - 0.82$. As $x\to-\infty$, $\lim_{x\to-\infty}9xe^{4x}=\lim_{x\to-\infty}\frac{9x}{e^{- 4x}}$. Using L'Hopital's rule (since it is in the $\frac{-\infty}{\infty}$ form), differentiating the numerator and denominator, we get $\lim_{x\to-\infty}\frac{9}{-4e^{-4x}} = 0$.

Answer:

Absolute maximum: $0$ Absolute minimum: $-0.82$ at $x =-\frac{1}{4}$