What is the extreme f(x)=-0.001x^2+6x-1900 ?

The extreme f(x)=-0.001x^2+6x-1900 is Maximum(3000,7100)

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extreme f(x)=-0.001x^2+6x-1900

extreme points

f (x) = - 0.001 x^{2} + 6 x - 1900

Maximum (3000, 7100)

Solution steps

f^{'} (x) = - 0.002 x + 6

Find intervals:

Increasing

: - \infty < x < 3000,

Decreasing

: 3000 < x < \infty

Plug

x = 3000

into

- 0.001 x^{2} + 6 x - 1900 : 7100

Maximum (3000, 7100)

h (x) = 5 (x - 1)^{\frac{2}{3}}, 0 \leq x \leq 2

f (x) = - 4 x^{3} + 6 x^{2} + 6

f (x, y) = \frac{5}{2} x^{2} + x y^{2} + 5 y^{2}

f (x) = 7 x^{3} - 7 x^{2} + 7

What is the extreme f(x)=-0.001x^2+6x-1900 ?
The extreme f(x)=-0.001x^2+6x-1900 is Maximum(3000,7100)