What is the extreme f(x)=x^3-4x^2-16x-5 ?

The extreme f(x)=x^3-4x^2-16x-5 is Maximum(-4/3 , 185/27),Minimum(4,-69)

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extreme f(x)=x^3-4x^2-16x-5

extreme points

f (x) = x^{3} - 4 x^{2} - 16 x - 5

Maximum (- \frac{4}{3}, \frac{185}{27}), Minimum (4, - 69)

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 - 5 : \frac{185}{27}

Maximum (- \frac{4}{3}, \frac{185}{27})

Plug

x = 4

into

x^{3} - 4 x^{2} - 16 x - 5 : - 69

Minimum (4, - 69)

Maximum (- \frac{4}{3}, \frac{185}{27}), Minimum (4, - 69)

f (x) = e^{x^{2} - 9}, - 3 \leq x \leq 3

x^{2} - 100 x

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

f (x) = - 2 x^{4} + 20 x^{2} - 18

f (x, y) = - x^{2} - y^{2} + 20 x + 20 y

What is the extreme f(x)=x^3-4x^2-16x-5 ?
The extreme f(x)=x^3-4x^2-16x-5 is Maximum(-4/3 , 185/27),Minimum(4,-69)