consider the function (f(x)=xe^{-9x}, 0leq xleq2). this function has an absolute minimum value equal to…

Explanation:

Step1: Find the derivative of $f(x)$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x$ and $v = e^{-9x}$. $u^\prime=1$ and $v^\prime=-9e^{-9x}$. So $f^\prime(x)=e^{-9x}-9xe^{-9x}=e^{-9x}(1 - 9x)$.

Step2: Find the critical points

Set $f^\prime(x)=0$. Since $e^{-9x}\gt0$ for all real $x$, we solve $1 - 9x = 0$. $1-9x = 0$ gives $x=\frac{1}{9}$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at $x = 0$, $x=\frac{1}{9}$, and $x = 2$. When $x = 0$, $f(0)=0\times e^{-9\times0}=0$. When $x=\frac{1}{9}$, $f(\frac{1}{9})=\frac{1}{9}e^{-9\times\frac{1}{9}}=\frac{1}{9e}$. When $x = 2$, $f(2)=2e^{-9\times2}=2e^{-18}$. Since $2e^{-18}\lt0\lt\frac{1}{9e}$, the absolute minimum value of $f(x)$ on $[0,2]$ is $2e^{-18}$ which occurs at $x = 2$, and the absolute maximum value is $\frac{1}{9e}$ which occurs at $x=\frac{1}{9}$.

Answer:

Absolute minimum value: $2e^{-18}$ $x$ - value for minimum: $2$ Absolute maximum value: $\frac{1}{9e}$ $x$ - value for maximum: $\frac{1}{9}$