Question

what is the maximum and minimum point of the function y=2x³-x²-4x+4​

Answer 1

$\huge\mathsf\pink{Answer}$

The minimum value is 0 and maximum value is -\frac{1}{2}

Step-by-step explanation:

Given : Function f(x)=x^4+2x^3-3x^2-4x+4

To find : The maximum and minimum value of function ?

Solution :

For maxima/ minima we find double derivate,

Function f(x)=x^4+2x^3-3x^2-4x+4

Derivate w.r.t x,

f'(x)=4x^3+6x^2-6x-4

For critical points put f'(x)=0,

4x^3+6x^2-6x-4=0

2x^3+3x^2-3x-2=0

2x^3-2x^2+5x^2-5x+2x-2=0

2x^2(x-1)+5x(x-1)+2(x-1)=0

(x-1)(2x^2+5x+2)=0

(x-1)(2x^2+4x+x+2)=0

(x-1)(x+2)(2x+1)=0

x=-2,-\frac{1}{2},1

Derivate again w.r.t x,

f''(x)=12x^2+12x-6

Check maxima/minima for each value of x,

1) f''(-2)=12(-2)^2+12(-2)-6

f''(-2)=18>0

It is minimum at x=-2.

The value is f(-2)=(-2)^4+2(-2)^3-3(-2)^2-4(-2)+4

f(-2)=16-16-12+8+4

f(-2)=0

2) f''(-\frac{1}{2})=12(-\frac{1}{2})^2+12(-\frac{1}{2})-6

f''(-\frac{1}{2})=-9

It is maximum

Answer 2

Explanation:

Important information regarding online test.

1. Once you use mobile version during test, you may find some questions incomplete, so you are instructed to use desktop version.

2.Try to submit before time. Thank you and all the best.