What is the extreme f(x)=4x^3-440x^2+10500x ?

The extreme f(x)=4x^3-440x^2+10500x is Maximum(15,72000),Minimum(175/3 ,-2450000/27)

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extreme f(x)=4x^3-440x^2+10500x

extreme

Maximum (15, 72000), Minimum (\frac{175}{3}, - \frac{2450000}{27})

Solution steps

f^{'} (x) = 12 x^{2} - 880 x + 10500

Find intervals:

Increasing

: - \infty < x < 15,

Decreasing

: 15 < x < \frac{175}{3},

Increasing

: \frac{175}{3} < x < \infty

Plug

x = 15

into

4 x^{3} - 440 x^{2} + 10500 x : 72000

Maximum (15, 72000)

Plug

x = \frac{175}{3}

into

4 x^{3} - 440 x^{2} + 10500 x : - \frac{2450000}{27}

Minimum (\frac{175}{3}, - \frac{2450000}{27})

Maximum (15, 72000), Minimum (\frac{175}{3}, - \frac{2450000}{27})

What is the extreme f(x)=4x^3-440x^2+10500x ?
The extreme f(x)=4x^3-440x^2+10500x is Maximum(15,72000),Minimum(175/3 ,-2450000/27)