Question

consider the function ( f(x)=x e^{-3 x}, 0 leq x leq 2 ).
this function has an absolute minimum value equal to:
which is attained at ( x=)
and an absolute maximum value equal to:
which is attained at ( x=)

Explanation:

Step1: Find the derivative of the function

Given \(f(x)=xe^{-3x}\), use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\), \(u^\prime=1\), \(v = e^{-3x}\), \(v^\prime=- 3e^{-3x}\). Then \(f^\prime(x)=e^{-3x}-3xe^{-3x}=e^{-3x}(1 - 3x)\)

Step2: Find the critical points

Set \(f^\prime(x) = 0\), since \(e^{-3x}>0\) for all \(x\), then \(1-3x = 0\), so \(x=\frac{1}{3}\)

Step3: Evaluate the function at critical and endpoints

• At \(x = 0\): \(f(0)=0\times e^{0}=0\)
• At \(x=\frac{1}{3}\): \(f(\frac{1}{3})=\frac{1}{3}e^{-3\times\frac{1}{3}}=\frac{1}{3e}\)
• At \(x = 2\): \(f(2)=2e^{-6}=\frac{2}{e^{6}}\)

Answer:

Absolute minimum value: \(0\) at \(x = 0\); Absolute maximum value: \(\frac{1}{3e}\) at \(x=\frac{1}{3}\)