this is similar to section 3.7 problem 8: for the function y = 5xe^{3x}, determine the absolute maximum and…

Explanation:

Step1: Find the derivative using product - rule

The product - rule states that if $y = uv$, where $u = 5x$ and $v=e^{3x}$, then $y^\prime=u^\prime v + uv^\prime$. $u^\prime = 5$ and $v^\prime=3e^{3x}$, so $y^\prime=5e^{3x}+5x\cdot3e^{3x}=5e^{3x}(1 + 3x)$.

Step2: Find the critical points

Set $y^\prime = 0$. Since $e^{3x}\gt0$ for all real $x$, we solve $1 + 3x=0$. $1+3x = 0$ gives $x=-\frac{1}{3}$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $y = 5xe^{3x}$ at $x=-1$, $x =-\frac{1}{3}$, and $x = 1$. When $x=-1$, $y=5\times(-1)\times e^{-3}=- \frac{5}{e^{3}}\approx - 0.25$. When $x =-\frac{1}{3}$, $y=5\times(-\frac{1}{3})\times e^{-1}=-\frac{5}{3e}\approx - 0.61$. When $x = 1$, $y=5\times1\times e^{3}=5e^{3}\approx100.43$.

Answer:

Absolute maximum = $100.43$ at $x = 1$. Absolute minimum = $-0.61$ at $x=-\frac{1}{3}$.