Answer to Question #188781 in Real Analysis for TUHIN SUBHRA DAS

Question

Determine the local minimum and local maximum value of the function f defined by f(x)=3-5x^3+5x^4-x^5

Answer 1

Solution:

$f(x)=3-5x^3+5x^4-x^5$

$\Rightarrow f'(x)=-15x^2+20x^3-5x^4$

Put $f'(x)=0$

$\Rightarrow -15x^2+20x^3-5x^4=0 \\\Rightarrow -5x^2(3-4x+x^2)=0$

$\\\Rightarrow -5x^2=0,(3-4x+x^2)=0$

$\\\Rightarrow x=0,(3-3x-x+x^2)=0 \\\Rightarrow(3-x)(1-x)=0 \\\Rightarrow x=1,x=3,x=0$

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

We have, $f'(x)>0$ when $1<x<3$ and $f'(x)<0$ when $x<0, 0<x<1,x>3$, so $x=3$ is a point of local maximum and $x=1$ is a point of local minimum.

So, put $x=3$ in $f(x)$.

$f(3)=3-5(3)^3+5(3)^4-(3)^5=30$

And put $x=1$ in $f(x)$.

$f(1)=3-5(1)^3+5(1)^4-(1)^5=2$

Hence, local maximum is 30 at x=3 and minimum is 2 at x=1.