Question

closed interval method: problem 3
(1 point)
results for this submission
2 of the answers are not correct.
consider the function $f(x)=xe^{-8x}$, $0\leq x\leq2$.
this function has an absolute minimum value equal to: $2e^{-16}$
which is attained at $x = 2$
and an absolute maximum value equal to: $\frac{1}{8e}$
which is attained at $x = \frac{1}{8}$
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Explanation:

Step1: Find the derivative of the function

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\), \(u^\prime=1\), \(v = e^{-8x}\), \(v^\prime=- 8e^{-8x}\).
So \(f^\prime(x)=e^{-8x}-8xe^{-8x}=e^{-8x}(1 - 8x)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), since \(e^{-8x}>0\) for all \(x\), then \(1 - 8x = 0\), which gives \(x=\frac{1}{8}\)

Step3: Evaluate the function at the critical point and endpoints

• When \(x = 0\), \(f(0)=0\times e^{-8\times0}=0\)
• When \(x=\frac{1}{8}\), \(f(\frac{1}{8})=\frac{1}{8}\times e^{-8\times\frac{1}{8}}=\frac{1}{8e}\)
• When \(x = 2\), \(f(2)=2\times e^{-8\times2}=2e^{-16}\approx2\times0.000001125 = 0.00000225\)

Answer:

The absolute minimum value is \(0\) which is attained at \(x = 0\), and the absolute maximum value is \(\frac{1}{8e}\) which is attained at \(x=\frac{1}{8}\)