consider the function\n\n$f(x)=x^{2}e^{x}$.\n\na) determine the intervals on which $f$ is concave up and…

consider the function\n\n$f(x)=x^{2}e^{x}$.\n\na) determine the intervals on which $f$ is concave up and concave down.\n\n$f$ is concave up on:\n\n$f$ is concave down on:\n\nb) 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 $(x,y)$).\n\n(separate multiple answers by commas.)\n\n(round to four decimal places as needed.)\n\nnote: when using interval notation in webwork, remember that:\n\n you use inf for $\\infty$ and -inf for $-\\infty$,\n\n and use u for the union symbol.\n\nenter dne if an answer does not exist.

consider the function\n\n$f(x)=x^{2}e^{x}$.\n\na) determine the intervals on which $f$ is concave up and concave down.\n\n$f$ is concave up on:\n\n$f$ is concave down on:\n\nb) 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 $(x,y)$).\n\n(separate multiple answers by commas.)\n\n(round to four decimal places as needed.)\n\nnote: when using interval notation in webwork, remember that:\n\n you use inf for $\\infty$ and -inf for $-\\infty$,\n\n and use u for the union symbol.\n\nenter dne if an answer does not exist.

Answer

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}<x<-2+\sqrt{2}), let (x=-2). Then (f^{\prime\prime}(-2)=e^{-2}((-2)^{2}+4\times(-2)+2)=e^{-2}(4-8 + 2)<0)
  • 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))