consider the function $f(x)=xe^{-5x}$, $0leq xleq2$. 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^{-5x}). (u^\prime=1), (v^\prime=- 5e^{-5x}) (f^\prime(x)=e^{-5x}-5xe^{-5x}=e^{-5x}(1 - 5x))

Step2: Find the critical points

Set (f^\prime(x)=0), since (e^{-5x}>0) for all (x), then (1 - 5x = 0), (x=\frac{1}{5}=0.2)

Step3: Evaluate (f(x)) at critical point and endpoints

• When (x = 0): (f(0)=0\times e^{-5\times0}=0)
• When (x=\frac{1}{5}): (f(\frac{1}{5})=\frac{1}{5}e^{-5\times\frac{1}{5}}=\frac{1}{5e})
• When (x = 2): (f(2)=2e^{-5\times2}=2e^{-10})

Compare the values: (2e^{-10}\approx2\times4.54\times10^{-5}), (\frac{1}{5e}\approx\frac{1}{5\times2.718}\approx0.0736), (0)

Answer:

Absolute minimum value: (0), attained at (x = 0) Absolute maximum value: (\frac{1}{5e}), attained at (x=\frac{1}{5})