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

Question #188781

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


1
Expert's answer
2021-05-07T11:44:20-0400

Solution:

f(x)=35x3+5x4x5f(x)=3-5x^3+5x^4-x^5

f(x)=15x2+20x35x4\Rightarrow f'(x)=-15x^2+20x^3-5x^4

Put f(x)=0f'(x)=0

15x2+20x35x4=05x2(34x+x2)=0\Rightarrow -15x^2+20x^3-5x^4=0 \\\Rightarrow -5x^2(3-4x+x^2)=0

5x2=0,(34x+x2)=0\\\Rightarrow -5x^2=0,(3-4x+x^2)=0

x=0,(33xx+x2)=0(3x)(1x)=0x=1,x=3,x=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)>0f'(x)>0 when 1<x<31<x<3 and f(x)<0f'(x)<0 when x<0,0<x<1,x>3x<0, 0<x<1,x>3, so x=3x=3 is a point of local maximum and x=1x=1 is a point of local minimum.

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

f(3)=35(3)3+5(3)4(3)5=30f(3)=3-5(3)^3+5(3)^4-(3)^5=30

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

f(1)=35(1)3+5(1)4(1)5=2f(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.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment