What is the extreme f(x)=-1/10 x^3+9x^2+500 ?

The extreme f(x)=-1/10 x^3+9x^2+500 is Minimum(0,500),Maximum(60,11300)

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extreme f(x)=-1/10 x^3+9x^2+500

extreme points

f (x) = - \frac{1}{10} x^{3} + 9 x^{2} + 500

Minimum (0, 500), Maximum (60, 11300)

Solution steps

f^{'} (x) = - \frac{3 x ^{2}}{10} + 18 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 60,

Decreasing

: 60 < x < \infty

Plug

x = 0

into

- \frac{1}{10} x^{3} + 9 x^{2} + 500 : 500

Minimum (0, 500)

Plug

x = 60

into

- \frac{1}{10} x^{3} + 9 x^{2} + 500 : 11300

Maximum (60, 11300)

Minimum (0, 500), Maximum (60, 11300)

f (x) = x (x - 2) e^{- 3 x}

f (x, y) = x^{3} - 4 x y + y^{2}

f (t) = - 4 t^{2} + 208 t + 120

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

f (x) = x (\frac{x ^{2}}{5} - 4)

What is the extreme f(x)=-1/10 x^3+9x^2+500 ?
The extreme f(x)=-1/10 x^3+9x^2+500 is Minimum(0,500),Maximum(60,11300)