Question

consider the function

$f(x)=x^{2}e^{x}$.

a) determine the intervals on which $f$ is concave up and concave down.

$f$ is concave up on:

$f$ is concave down on:

b) based on your answer to part (a), determine the inflection points of $f$. each point should be entered as an ordered pair (that is, in the form $(x,y)$).

(separate multiple answers by commas.)

(round to four decimal places as needed.)

note: when using interval notation in webwork, remember that:

you use inf for $\infty$ and -inf for $-\infty$,

and use u for the union symbol.

enter dne if an answer does not exist.

Explanation:

Step1: Find the first derivative

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

Step2: Find the second derivative

Again using the product rule. Let \(u = x^{2}+2x\), \(u^\prime = 2x + 2\), \(v=e^{x}\), \(v^\prime=e^{x}\)
\(f^{\prime\prime}(x)=e^{x}(x^{2}+2x)+e^{x}(2x + 2)=e^{x}(x^{2}+4x + 2)\)

Step3: Find the critical points of \(f^{\prime\prime}(x)\)

Set \(f^{\prime\prime}(x)=0\), since \(e^{x}>0\) for all \(x\in R\), we solve \(x^{2}+4x + 2=0\)
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b = 4\), \(c=2\)
\(x=\frac{-4\pm\sqrt{16 - 8}}{2}=\frac{-4\pm2\sqrt{2}}{2}=-2\pm\sqrt{2}\)

Step4: Test the intervals for concavity

• For \(x<-2-\sqrt{2}\), let \(x=-4\). Then \(f^{\prime\prime}(-4)=e^{-4}((-4)^{2}+4\times(-4)+2)=e^{-4}(16 - 16+2)>0\)
• For \(-2-\sqrt{2}
• For \(x>-2+\sqrt{2}\), let \(x=0\). Then \(f^{\prime\prime}(0)=e^{0}(0^{2}+4\times0+2)=2>0\)

Answer:

a) \(f\) is concave up on: \((-\infty,-2-\sqrt{2})\cup(-2 + \sqrt{2},\infty)\)
\(f\) is concave down on: \((-2-\sqrt{2},-2+\sqrt{2})\)
b) Inflection points: \((-2-\sqrt{2},(-2-\sqrt{2})^{2}e^{-2-\sqrt{2}}),(-2+\sqrt{2},(-2+\sqrt{2})^{2}e^{-2+\sqrt{2}})\)
Calculating the \(y -\)values:
\((-2-\sqrt{2})^{2}=4 + 4\sqrt{2}+2=6 + 4\sqrt{2}\approx6+4\times1.4142 = 11.6568\), \(e^{-2-\sqrt{2}}\approx e^{-3.4142}\approx0.0339\), \(y_1\approx11.6568\times0.0339\approx0.3952\)
\((-2+\sqrt{2})^{2}=4-4\sqrt{2}+2=6 - 4\sqrt{2}\approx6-4\times1.4142=0.3432\), \(e^{-2+\sqrt{2}}\approx e^{-0.5858}\approx0.5564\), \(y_2\approx0.3432\times0.5564\approx0.1910\)

So the inflection points are \((-3.4142,0.3952),(-0.5858,0.1910)\)