Question

find the local and global maximum and minimum value of the function f (x)=4x^3-6x^2-9x+1 on the interval [-1,2]​

Answer 1

Answer:

put x+1=0 or x=-1. f(-1)is the remainder. now, f(-1)=(-1)^3+(-1)^2-3(-1)+2=-1+1+3+2=5therefore 5 is remainder

Answer 2

Answer:

The global maximum is 15 and minimum is -9.

And local maxima is 24 and local minima is -24.

and $f'' (x)=24 > 0$ therefore $x=\dfrac{3}{2}$ is the local minima.

Step-by-step explanation:

Given the equation, $f (x)=4x^3-6x^2-9x+1$

Differentiating with respect to x,

$f' (x)=12x^2-12x-9$                      ...(1)

To solve for x, take

$12x^2-12x-9=0\implies 4x^2-4x-3=0$

$\implies 4x^2+2x-6x-3=0$

$\implies 2x(2x+1)-3(2x+1)=0$

$\implies (2x-3)(2x+1)=0$

Therefore, the values $x=-\dfrac{1}{2} ~~or~~\dfrac{3}{2}$  are the critical values of the function.

Differentiating equation (1) again with respect to x,

$f'' (x)=24x-12$                    

Substituting the values of x,

$f'' (x)=24\big(-\frac{1}{2} \big)-12=-24$

$f'' (x)=24\big(\frac{3}{2} \big)-12=24$

Therefore, $f'' (x)=-24 < 0$ therefore $x=-\dfrac{1}{2}$ is the local maxima.

and $f'' (x)=24 > 0$ therefore $x=\dfrac{3}{2}$ is the local minima.

For global maxima, the interval is given,  [-1, 2]​

Then the required x values are $[-1, \frac{3}{2}, 2]$

Substituting these values in equation (1),

$f' (x)=12(-1)^2-12(-1)-9=12-12-9=-9$

$f' (x)=12(2)^2-12(2)-9=48-24-9=15$

$f' (x)=12(\frac{3}{2} )^2-12(\frac{3}{2} )-9=0$

Therefore, the global maximum is 15 for $x = 2$ and global minimum is for $x = - 9$.