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

The extreme f(x)=t^3-9t^2 is Maximum(0,0),Minimum(6,-108)

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

extreme points

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

Maximum (0, 0), Minimum (6, - 108)

Solution steps

f^{'} (t) = 3 t^{2} - 18 t

Find intervals:

Increasing

: - \infty < t < 0,

Decreasing

: 0 < t < 6,

Increasing

: 6 < t < \infty

Plug

t = 0

into

t^{3} - 9 t^{2} : 0

Maximum (0, 0)

Plug

t = 6

into

t^{3} - 9 t^{2} : - 108

Minimum (6, - 108)

Maximum (0, 0), Minimum (6, - 108)

y = \frac{( x ( z + 2 ) )}{7}

- \frac{x ^{4}}{4} + \frac{x ^{2}}{4}

f (x, y) = - 9 x^{2} - 2 y^{2} + 5 x - 9 y + 10

f (x) = \frac{18}{x ^{2} + 2}

f (x) = - 3 x + ln (x), (0, 3)

What is the extreme f(x)=t^3-9t^2 ?
The extreme f(x)=t^3-9t^2 is Maximum(0,0),Minimum(6,-108)