What is the extreme f(x)=110.2x-0.4x^2-20.2-2000 ?

The extreme f(x)=110.2x-0.4x^2-20.2-2000 is Maximum(551/4 ,5569.825)

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extreme f(x)=110.2x-0.4x^2-20.2-2000

extreme points

f (x) = 110.2 x - 0.4 x^{2} - 20.2 - 2000

Maximum (\frac{551}{4}, 5569.825)

Solution steps

f^{'} (x) = - 0.8 x + 110.2

Find intervals:

Increasing

: - \infty < x < \frac{551}{4},

Decreasing

: \frac{551}{4} < x < \infty

Plug

x = \frac{551}{4}

into

110.2 x - 0.4 x^{2} - 20.2 - 2000 : 5569.825

Maximum (\frac{551}{4}, 5569.825)

f (x) = \frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36

y = 4 x^{3} - 8 x

f (x) = \frac{7}{4 - 3 sin ^{2} ( x )}

x^{2} - 5, 1 < x < 7

(x - 7) (x - 5)^{2} (x - 3) (x - 1)

What is the extreme f(x)=110.2x-0.4x^2-20.2-2000 ?
The extreme f(x)=110.2x-0.4x^2-20.2-2000 is Maximum(551/4 ,5569.825)