Question

For Problem 1, find the exact values of the critical points of f. 1. f(x) = x3 + 3x2 – 24.c + 5

Answer 1

$f\left ( x \right )=x^{3}+3x^{2}-24x+5$

To find critical points of f derive the function first and equates to zero

S' () = (23 + 3x² – 24x + 5)

$f'\left ( x \right )=\left (3x^{3-1}+3(2)x^{2-1}-24(1)+0 \right )$         Apply the formula $\frac{d}{dx}\left ( x^{n} \right )=nx^{n-1}$

$f'\left ( x \right )=\left (3x^{2}+6x^{1}-24\right )$

To find critical points $f'(x)=0$

$\left (3x^{2}+6x-24\right )=0$

$3\left (x^{2}+2x-8\right )=0$

(2? + 2.1 – 8) = 0

$\left (x^{2}+4x-2x-8\right )=0$                Apply Sum and product of zeros

$x\left ( x+4 \right )-2\left ( x+4 \right )=0$           Common same terms out side

$\left ( x+4 \right ) \left ( x-2 \right )=0$                     Common$\left ( x+4 \right )$ out side

$\left ( x+4 \right ) =0 or \left ( x-2 \right )=0$

$f(-4)=\left ( -4 \right )^{3}+3\left ( -4 \right )^{2}-24\left ( -4 \right )+4$

$f(-4)=-64+3\left ( 16 \right )+96+4$

   $f(-4)=-64+48+96+4$

$f(-4)=-64+148$

$f(-4)=-84$

$f(2)=2^{3}+3\left ( 2 \right )^{2}-24\left ( 2 \right )+5$

$f(2)=8+3\left ( 4 \right )-48+5$

$f(2)=8+12-48+5$

$f(2)=25-48$

$f(2)=-23$              

Critical points are $\left ( -4,-84 \right )$ and $\left ( 2,-23 \right )$