Question

find the maximum and minimum valuey
of function :
f(x) = x⁴ + 2x³ – 3x² - 4x + 4​

Answer 1

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

$f''(-\frac{1}{2})=-9<0$

It is maximum at $x=-\frac{1}{2}$

The value is $f(-\frac{1}{2})=(-\frac{1}{2})^4+2(-\frac{1}{2})^3-3(-\frac{1}{2})^2-4(-\frac{1}{2})+4$

$f(-\frac{1}{2})=5.0625$

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

$f''(1)=18>0$

It is minimum at $x=-\frac{1}{2}$

The value is $f(1)=(1)^4+2(1)^3-3(1)^2-4(1)+4$

$f(1)=0$

#Learn more

F(x)=x¹/³(x+3)²/³ x ∈ R⁺,Find the maximum and minimum value of the given function.

https://brainly.in/question/5598229

Answer 2

Answer:

Step-by-step explanation: