Question

hw 16 - first derivative test section 3.2: problem 10
(1 point)
consider the function
$f(x)=6 x^{2} e^{-x}$.

1. $f$ is increasing on the intervals
2. $f$ is decreasing on the intervals
3. the relative maxima of $f$ occur at $x=$
4. the relative minima of $f$ occur at $x=$

notes: in the last two, your answer should be a comma separated list of $x$ values or the word
one\.
note: you can earn partial credit on this problem.
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Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 6x^{2}\), \(u^\prime=12x\), \(v = e^{-x}\), \(v^\prime=-e^{-x}\).
\(f^\prime(x)=12xe^{-x}-6x^{2}e^{-x}=6xe^{-x}(2 - x)\)

Step2: Find critical points

Set \(f^\prime(x)=0\), \(6xe^{-x}(2 - x)=0\). Since \(e^{-x}>0\) for all \(x\), then \(x = 0\) or \(x = 2\)

Step3: Test intervals

• For \(x<0\), let \(x=-1\), \(f^\prime(-1)=6\times(-1)\times e^{1}(2+1)=-18e<0\)
• For \(0
• For \(x>2\), let \(x=3\), \(f^\prime(3)=6\times3\times e^{-3}(2 - 3)=- \frac{18}{e^{3}}<0\)

Step4: Determine increasing/decreasing intervals and extrema

• \(f(x)\) is increasing when \(f^\prime(x)>0\), so on the interval \((0,2)\)
• \(f(x)\) is decreasing when \(f^\prime(x)<0\), so on the intervals \((-\infty,0)\cup(2,\infty)\)
• Since \(f(x)\) changes from decreasing to increasing at \(x = 0\), \(x = 0\) is a relative minimum.
• Since \(f(x)\) changes from increasing to decreasing at \(x = 2\), \(x = 2\) is a relative maximum.

Answer:

1. \((0,2)\)
2. \((-\infty,0)\cup(2,\infty)\)
3. \(2\)
4. \(0\)