What is the extreme f(x)=x^3-6x^2+16 ?

The extreme f(x)=x^3-6x^2+16 is Maximum(0,16),Minimum(4,-16)

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extreme f(x)=x^3-6x^2+16

extreme points

f (x) = x^{3} - 6 x^{2} + 16

Maximum (0, 16), Minimum (4, - 16)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

x^{3} - 6 x^{2} + 16 : 16

Maximum (0, 16)

Plug

x = 4

into

x^{3} - 6 x^{2} + 16 : - 16

Minimum (4, - 16)

Maximum (0, 16), Minimum (4, - 16)

f (x) = \frac{2}{3} x^{3} + 1

f (x) = \frac{( x + 3 )}{x + 9}

\frac{- 4 - 3 x}{7 x - 5}

f (x) = x^{3} - 3 x^{2} + 2 x + 1

f (x) = - x^{2}

What is the extreme f(x)=x^3-6x^2+16 ?
The extreme f(x)=x^3-6x^2+16 is Maximum(0,16),Minimum(4,-16)