What is the extreme f(x)=4x^3-46x^2+120x ?

The extreme f(x)=4x^3-46x^2+120x is Maximum(5/3 , 2450/27),Minimum(6,-72)

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extreme f(x)=4x^3-46x^2+120x

extreme

Maximum (\frac{5}{3}, \frac{2450}{27}), Minimum (6, - 72)

Solution steps

f^{'} (x) = 12 x^{2} - 92 x + 120

Find intervals:

Increasing

: - \infty < x < \frac{5}{3},

Decreasing

: \frac{5}{3} < x < 6,

Increasing

: 6 < x < \infty

Plug

x = \frac{5}{3}

into

4 x^{3} - 46 x^{2} + 120 x : \frac{2450}{27}

Maximum (\frac{5}{3}, \frac{2450}{27})

Plug

x = 6

into

4 x^{3} - 46 x^{2} + 120 x : - 72

Minimum (6, - 72)

Maximum (\frac{5}{3}, \frac{2450}{27}), Minimum (6, - 72)

What is the extreme f(x)=4x^3-46x^2+120x ?
The extreme f(x)=4x^3-46x^2+120x is Maximum(5/3 , 2450/27),Minimum(6,-72)