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Find the third derivative of $f(x)=xe^{(x^3)}$. Along with the graph of the third derivative show the points where the third derivative is $0$.

This is a problem that I am supposed to solve in Wolfram Mathematica. I have begun with finding the derivative using D[[f[x],{x,3}] but I am having trouble finding the points and showing them on the graph. In the presentations from my school it says that I am supposed to use the function Last[Last[FindRoot[f'[x], {x,0}]]], but I don't understand why I should use it because it's not really showing anything on my graph.

By the way, {x,0} in the FindRoot[] function is supposed to be find the solutions for the equation in the area around 0 or ..? I'm guessing that since the graph of the derivative looks like this:

enter image description here

The whole code that I have written for this problem is this: GRAFIK[x_] = x*E^(x^3)

r1[x_] = D[x*E^(x^3), {x, 3}]

Simplify[3 (6 E^x^3 x + 9 E^x^3 x^4) + x (6 E^x^3 + 54 E^x^3 x^3 + 27 E^x^3 x^6)]

a = Plot[r1[x], {x, -1.5, 1}]

rese1 = Last[Last[FindRoot[r1[x] == 0, {x, 0}]]]

rese2 = Last[Last[FindRoot[r1[x] == 0, {x, -1.5}]]]

rese3 = Last[Last[FindRoot[r1[x] == 0, {x, -0.5}]]]

b = ListPlot[{rese1, GRAFIK[rese1]}, PlotStyle -> PointSize[0.02]];

c = ListPlot[{rese2, GRAFIK[rese2]}, PlotStyle -> PointSize[0.02]];

d = ListPlot[{rese3, GRAFIK[rese3]}, PlotStyle -> PointSize[0.02]];

Show[{a, b}, {a, c}, {a, d}]

1 Answers 1

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Using Wolfram Mathematica like you asked, you can easily find that the this derivate is

$$f^{(3)}\colon x\mapsto (27x^7+81x^4+24x)e^{x^3}.$$

Then you know since $e^X>0$ for all $X$ that $f^{(3)}(x)=0$ if, and only if,

$$27x^7+81x^4+24x=0.$$

Know you just have to solve this equation using Wolfram Mathematica, and you get three solutions:

$$\left\{0,-\frac 2{\sqrt[3] 3},-\frac 1{\sqrt[3] 3}\right\}.$$

  • 0
    This reduces to a cubic equation in $x^3$ plus $x=0$2017-01-21
  • 0
    I've tried that as well. I found the solutions to the equation using FindRoot and then tried to show them on the graph using ListPlot but I don't know, nothing worked.2017-01-21