consider the function $f(x)=xe^{-9x}, 0leq xleq 2$. this function has an absolute minimum value equal to…

Explanation:

Step1: Find the derivative

Use 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 critical points

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

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)$ at $x = 0$, $x=\frac{1}{9}$, and $x = 2$. $f(0)=0\times e^{-9\times0}=0$; $f(\frac{1}{9})=\frac{1}{9}\times e^{-9\times\frac{1}{9}}=\frac{1}{9e}$; $f(2)=2\times e^{-9\times2}=2e^{-18}$. Since $2e^{-18}\approx2\times6.74\times10^{-8}\approx1.35\times10^{-7}$ and $0\lt2e^{-18}\lt\frac{1}{9e}$, the absolute minimum value is $0$ at $x = 0$.

Answer:

The absolute minimum value is $0$ which is attained at $x = 0$.