Question

consider the function ( f(x)=x e^{-2 x}, quad 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=).
(round to three decimal places as needed.)

Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\), \(u^\prime=1\), \(v = e^{-2x}\), \(v^\prime=- 2e^{-2x}\).
\(f^\prime(x)=e^{-2x}-2xe^{-2x}=e^{-2x}(1 - 2x)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\). Since \(e^{-2x}>0\) for all \(x\), then \(1 - 2x = 0\), so \(x=\frac{1}{2}\)

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

• When \(x = 0\): \(f(0)=0\times e^{-2\times0}=0\)
• When \(x=\frac{1}{2}\): \(f(\frac{1}{2})=\frac{1}{2}e^{-2\times\frac{1}{2}}=\frac{1}{2e}\approx0.184\)
• When \(x = 2\): \(f(2)=2e^{-2\times2}=2e^{-4}\approx0.037\)

Answer:

The absolute minimum value is \(0\) which is attained at \(x = 0\), and the absolute maximum value is \(\frac{1}{2e}\approx0.184\) which is attained at \(x=\frac{1}{2}\)