Question

Locate the absolute maximum and minimum for each of the following functions and justify your responses. (Show all work)

Question 1: 𝑓(𝑥) = Cube root of x^2 − 𝑥 on the interval [−1,1] 

Question 2: g(x)=xe^2x on the interval [-2,0]

First and Second Derivative Analysis :For each of the following functions, respond to the given prompts. Show all work that leads to your responses:

Given 𝑓(𝑥) = 3𝑥 3 − 18𝑥 2 − 45𝑥 + 10

a. On what interval(s), is 𝑓(𝑥) increasing? Justify. show all work

b. At what value(s) of 𝑥 does 𝑓(𝑥) have a relative minimum? Justify. show all work

c. On what interval(s), is 𝑓(𝑥) decreasing and concave up? Justify. show all work

Answer 1

The absolute maximum and minimum

$\mathrm{Suppose\:that\:}x=c\mathrm{\:is\:a\:critical\:point\:of\:}f\left(x\right)\mathrm{\:then,\:}$

\mathrm{If\:}f\:'\left(x\right)>0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:maximum.}

\mathrm{If\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)>\:0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:minimum.}

\mathrm{If\:}f\:'\left(x\right)\mathrm{\:is\:the\:same\:sign\:on\:both\:sides\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:neither\:a\:local\:maximum\:nor\:a\:local\:minimum.}

$f'(x)$ gives stationary points. Hence differentiation with respect to x

1) Applying chain rule $f'(x)=\frac{1}{3\left(x^2-x\right)^{\frac{2}{3}}}\frac{d}{dx}\left(x^2-x\right)=\frac{1}{3\left(x^2-x\right)^{\frac{2}{3}}}\left(2x-1\right)$

$f'(x)=\frac{2x-1}{3\left(x^2-x\right)^{\frac{2}{3}}}$

Hence $(x,f(x))=(−1,-2\sqrt3)$ for absolute maximum

Hence $(x,f(x))=(−1,-\frac{\sqrt3}{4})$ for absolute minimum

2) First derivative for $3x^3−18x^2−\:45x+10$

$\implies f'(x)=\frac{d}{dx}\left(3x^3\right)-\frac{d}{dx}\left(18x^2\right)-\frac{d}{dx}\left(45x\right)+\frac{d}{dx}\left(10\right)$

$\implies f'(x)=9x^2-36x-45$

Second derivative for $3x^3−18x^2−\:45x+10$

$\implies f''(x)=\frac{d}{dx}(9x^2)-\frac{d}{dx}(36x)-\frac{d}{dx}(45)$

$\implies f''(x)=18x-36$

a. Increasing interval(s) of 𝑓(𝑥) \mathrm{If\:}f\:'\left(x\right)>0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:maximum.}

Hence $f'(x)>0$ is the increasing intervals $\implies$ $(-\infty <x<-1) and (5<x<\infty )$

b. A relative minimum is when $f'\left(x\right)=9x^2−36x−45=0$

$x_{1,\:2}=\frac{-\left(-36\right)\pm \sqrt{\left(-36\right)^2-4\cdot \:9\left(-45\right)}}{2\cdot \:9}$

$x_{1,\:2}=\frac{-\left(-36\right)\pm \:54}{2\cdot \:9}$

$x_1=\frac{-\left(-36\right)+54}{2\cdot \:9},\:x_2=\frac{-\left(-36\right)-54}{2\cdot \:9}$

$x=5,\:x=-1$

$\mathrm{Minimum}\left(5,\:-290\right)$

At x=5, 𝑓(𝑥) have a relative minimum

c. Decreasing interval(s) of 𝑓(𝑥) \mathrm{If\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)>0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:maximum.}

Hence $f'(x)<0$ is the decreasing interval $\implies (-1<x<5)$