What is the extreme f(x)=4x^3-80x^2+400x ?

The extreme f(x)=4x^3-80x^2+400x is Maximum(10/3 , 16000/27),Minimum(10,0)

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extreme f(x)=4x^3-80x^2+400x

extreme points

f (x) = 4 x^{3} - 80 x^{2} + 400 x

Maximum (\frac{10}{3}, \frac{16000}{27}), Minimum (10, 0)

Solution steps

f^{'} (x) = 12 x^{2} - 160 x + 400

Find intervals:

Increasing

: - \infty < x < \frac{10}{3},

Decreasing

: \frac{10}{3} < x < 10,

Increasing

: 10 < x < \infty

Plug

x = \frac{10}{3}

into

4 x^{3} - 80 x^{2} + 400 x : \frac{16000}{27}

Maximum (\frac{10}{3}, \frac{16000}{27})

Plug

x = 10

into

4 x^{3} - 80 x^{2} + 400 x : 0

Minimum (10, 0)

Maximum (\frac{10}{3}, \frac{16000}{27}), Minimum (10, 0)

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

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

y = x

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

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

What is the extreme f(x)=4x^3-80x^2+400x ?
The extreme f(x)=4x^3-80x^2+400x is Maximum(10/3 , 16000/27),Minimum(10,0)