What is the extreme f(x)=x^3-36x^2+324x-3280.5 ?

The extreme f(x)=x^3-36x^2+324x-3280.5 is Maximum(6,-2416.5),Minimum(18,-3280.5)

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extreme f(x)=x^3-36x^2+324x-3280.5

extreme points

f (x) = x^{3} - 36 x^{2} + 324 x - 3280.5

Maximum (6, - 2416.5), Minimum (18, - 3280.5)

Solution steps

f^{'} (x) = 3 x^{2} - 72 x + 324

Find intervals:

Increasing

: - \infty < x < 6,

Decreasing

: 6 < x < 18,

Increasing

: 18 < x < \infty

Plug

x = 6

into

x^{3} - 36 x^{2} + 324 x - 3280.5 : - 2416.5

Maximum (6, - 2416.5)

Plug

x = 18

into

x^{3} - 36 x^{2} + 324 x - 3280.5 : - 3280.5

Minimum (18, - 3280.5)

Maximum (6, - 2416.5), Minimum (18, - 3280.5)

h (v, t) = v t - 5 t^{2}

f (x) = \frac{( 3 e ^{x} )}{x ^{3}}

f (x) = \frac{6}{5} x^{\frac{5}{2}} - \frac{4}{3} x^{\frac{3}{2}} + 1

f (x) = 43 x + 8 cos (x)

f (x, y) = x \cdot e^{- 2 x^{2} - 2 y^{2}}

What is the extreme f(x)=x^3-36x^2+324x-3280.5 ?
The extreme f(x)=x^3-36x^2+324x-3280.5 is Maximum(6,-2416.5),Minimum(18,-3280.5)