What is the extreme f(x)=x^3+8x^2+16x+3 ?

The extreme f(x)=x^3+8x^2+16x+3 is Maximum(-4,3),Minimum(-4/3 ,-175/27)

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extreme f(x)=x^3+8x^2+16x+3

extreme

Maximum (- 4, 3), Minimum (- \frac{4}{3}, - \frac{175}{27})

Solution steps

f^{'} (x) = 3 x^{2} + 16 x + 16

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < - \frac{4}{3},

Increasing

: - \frac{4}{3} < x < \infty

Plug

x = - 4

into

x^{3} + 8 x^{2} + 16 x + 3 : 3

Maximum (- 4, 3)

Plug

x = - \frac{4}{3}

into

x^{3} + 8 x^{2} + 16 x + 3 : - \frac{175}{27}

Minimum (- \frac{4}{3}, - \frac{175}{27})

Maximum (- 4, 3), Minimum (- \frac{4}{3}, - \frac{175}{27})

What is the extreme f(x)=x^3+8x^2+16x+3 ?
The extreme f(x)=x^3+8x^2+16x+3 is Maximum(-4,3),Minimum(-4/3 ,-175/27)