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

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

x=-4 or x=2

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 )

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