Question

the graph of y=(x^2-4)^4(x^2+1)^ 5 is shown to the right. Find the coordinates of the local maximum points and minimum points

Answer 1

Step 1

Given;

$y=(x^2-4)^4(x^2+1)^5$

Required; To find the coordinates of the local minima and maxima

Step 2

Find the local minima and maxima

$\begin{gathered} \mathrm{Suppose\:that\:}x=c\mathrm{\:is\:a\:critical\:point\:of\:}f\left(x\right)\mathrm{\:then,\:} \\ \mathrm{If\:}f\:'\left(x\right)>0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:maximum.} \\ \mathrm{If\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)>\:0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:minimum.} \\ \mathrm{If\:}f\:'\left(x\right)\mathrm{\:is\:the\:same\:sign\:on\:both\:sides\:of\:}x=c\mathrm{\:then\:}x=c \\ \mathrm{\:is\:neither\:a\:local\:maximum\:nor\:a\:local\:minimum.} \end{gathered}$

Therefore; f'(x) is given as;

$f^{\prime}(x)=2x(x^2-4)^3(x^2+1)^4[9x^2-16]$

Step 3

Find the increasing and decreasing intervals from the graph

$\begin{gathered} Decreasing;-\inftyPlugin x=-2 into y[tex]\begin{gathered} \mathrm{Minimum}\left(-2,\:0\right) \\ \end{gathered}$

Plugin -4/3 into y

$\mathrm{Maximum}\left(-\frac{4}{3},\:\frac{5^{14}\cdot \:256}{387420489}\right)$

Plugin x=0 into y

$\mathrm{Minimum}\left(0,\:256\right)$

Plugin x=4/3

$\mathrm{Maximum}\left(\frac{4}{3},\:\frac{5^{14}\cdot \:256}{387420489}\right)$

Plugging x=2 into y

$\mathrm{Minimum}\left(2,\:0\right)$

Answer; The maximum points are;

$\begin{gathered} (\frac{4}{3},\frac{5^{14}\times256}{387420489})\text{ or \lparen1.33},4033.09) \\ \left(-\frac{4}{3},\:\frac{5^{14}\cdot\:256}{387420489}\right)or\text{ \lparen-1.33},\text{ 4033.09\rparen} \end{gathered}$

The minimum points are ;

$\begin{gathered} \left(-2,\:0\right) \\ \left(0,\:256\right) \\ \left(2,\:0\right) \end{gathered}$