What is the extreme f(x)=x^3-4x^2-16x+9 ?

The extreme f(x)=x^3-4x^2-16x+9 is Maximum(-4/3 , 563/27),Minimum(4,-55)

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extreme f(x)=x^3-4x^2-16x+9

extreme points

f (x) = x^{3} - 4 x^{2} - 16 x + 9

Maximum (- \frac{4}{3}, \frac{563}{27}), Minimum (4, - 55)

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 + 9 : \frac{563}{27}

Maximum (- \frac{4}{3}, \frac{563}{27})

Plug

x = 4

into

x^{3} - 4 x^{2} - 16 x + 9 : - 55

Minimum (4, - 55)

Maximum (- \frac{4}{3}, \frac{563}{27}), Minimum (4, - 55)

f (x) = 2 x - 4 x

f (x) = x + (1 - x)

y = \frac{x ^{2}}{( x - 2 ) ^{2}}

- 2 x - 1

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

What is the extreme f(x)=x^3-4x^2-16x+9 ?
The extreme f(x)=x^3-4x^2-16x+9 is Maximum(-4/3 , 563/27),Minimum(4,-55)