Question

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

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 \(x=\frac{1}{5}\)