Question

closed interval method: problem 3
(1 point)
consider the function $f(x)=x e^{-8 x}, quad 0 leq x leq 2$.
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 $x=$
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Explanation:

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

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

Step2: Find the critical points

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

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

• When \(x = 0\): \(f(0)=0\times e^{-8\times0}=0\)
• When \(x=\frac{1}{8}\): \(f(\frac{1}{8})=\frac{1}{8}e^{-8\times\frac{1}{8}}=\frac{1}{8e}\)
• When \(x = 2\): \(f(2)=2e^{-8\times2}=2e^{-16}\)

Since \(2e^{-16}\approx2\times\frac{1}{e^{16}}\approx0\) (because \(e^{16}\gg1\))

Answer:

The absolute minimum value is \(2e^{-16}\) which is attained at \(x = 2\)
The absolute maximum value is \(\frac{1}{8e}\) which is attained at \(x=\frac{1}{8}\)