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webwork 4 - topics 8 - 9: problem 2
(1 point)
consider the function ( f(x)=x e^{-7 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=)

Explanation:

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

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\) and \(v=e^{-7x}\).
\(u^\prime=1\), \(v^\prime=- 7e^{-7x}\)
\(f^\prime(x)=e^{-7x}-7xe^{-7x}=e^{-7x}(1 - 7x)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), since \(e^{-7x}>0\) for all \(x\), then \(1 - 7x = 0\), \(x=\frac{1}{7}\)

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

• At \(x = 0\): \(f(0)=0\times e^{-7\times0}=0\)
• At \(x=\frac{1}{7}\): \(f(\frac{1}{7})=\frac{1}{7}e^{-7\times\frac{1}{7}}=\frac{1}{7e}\)
• At \(x = 2\): \(f(2)=2e^{-7\times2}=2e^{-14}\)

Since \(2e^{-14}\approx2\times1.234\times10^{-6}\approx2.468\times10^{-6}\), \(\frac{1}{7e}\approx\frac{1}{7\times2.718}\approx0.052\)

Answer:

The absolute minimum value is \(2e^{-14}\) which is attained at \(x = 2\).
The absolute maximum value is \(\frac{1}{7e}\) which is attained at \(x=\frac{1}{7}\)