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

The extreme f(x)=x^3-3x^2+9 is Maximum(0,9),Minimum(2,5)

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

extreme points

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

Maximum (0, 9), Minimum (2, 5)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, 9)

Plug

x = 2

into

x^{3} - 3 x^{2} + 9 : 5

Minimum (2, 5)

Maximum (0, 9), Minimum (2, 5)

- \frac{10}{x ^{2}}

f (x) = (x + 4)^{\frac{2}{7}}

y = x^{2} + x

f (x) = \frac{\frac{3}{2}}{2 \frac{5}{2} + \frac{3}{2} x}

x - 5

What is the extreme f(x)=x^3-3x^2+9 ?
The extreme f(x)=x^3-3x^2+9 is Maximum(0,9),Minimum(2,5)