What is the extreme f(x)=x^3-9x^2+15 ?

The extreme f(x)=x^3-9x^2+15 is Maximum(0,15),Minimum(6,-93)

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extreme f(x)=x^3-9x^2+15

extreme points

f (x) = x^{3} - 9 x^{2} + 15

Maximum (0, 15), Minimum (6, - 93)

Solution steps

f^{'} (x) = 3 x^{2} - 18 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

x^{3} - 9 x^{2} + 15 : 15

Maximum (0, 15)

Plug

x = 6

into

x^{3} - 9 x^{2} + 15 : - 93

Minimum (6, - 93)

Maximum (0, 15), Minimum (6, - 93)

f (x) = 2 x^{3} - 39 x^{2} + 240 x + 8

f (x) = \frac{9 x ^{2} + x}{x ^{2} + 4}

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

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

f (x) = \frac{( - x ^{2} + 8 )}{( 2 x ^{2} - 3 )}

What is the extreme f(x)=x^3-9x^2+15 ?
The extreme f(x)=x^3-9x^2+15 is Maximum(0,15),Minimum(6,-93)