What is the extreme f(x)=x^3-4x^2-16x-3 ?

The extreme f(x)=x^3-4x^2-16x-3 is Maximum(-4/3 , 239/27),Minimum(4,-67)

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extreme f(x)=x^3-4x^2-16x-3

extreme points

f (x) = x^{3} - 4 x^{2} - 16 x - 3

Maximum (- \frac{4}{3}, \frac{239}{27}), Minimum (4, - 67)

Solution steps

f^{'} (x) = 3 x^{2} - 8 x - 16

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 4 < x < \infty

Plug

x = - \frac{4}{3}

into

x^{3} - 4 x^{2} - 16 x - 3 : \frac{239}{27}

Maximum (- \frac{4}{3}, \frac{239}{27})

Plug

x = 4

into

x^{3} - 4 x^{2} - 16 x - 3 : - 67

Minimum (4, - 67)

Maximum (- \frac{4}{3}, \frac{239}{27}), Minimum (4, - 67)

f (x) = \frac{1}{( x + 1 ) ^{2}} + 2

f (x) = - x^{2} - 1

\frac{2 x}{9 - x ^{2}}

8 x - 4 y = 16

f (x) = \frac{2 x + 3}{x ^{3}}

What is the extreme f(x)=x^3-4x^2-16x-3 ?
The extreme f(x)=x^3-4x^2-16x-3 is Maximum(-4/3 , 239/27),Minimum(4,-67)