What is the extreme f(x)=-0.05x^2+300x+2000 ?

The extreme f(x)=-0.05x^2+300x+2000 is Maximum(3000,452000)

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extreme f(x)=-0.05x^2+300x+2000

extreme points

f (x) = - 0.05 x^{2} + 300 x + 2000

Maximum (3000, 452000)

Solution steps

f^{'} (x) = - 0.1 x + 300

Find intervals:

Increasing

: - \infty < x < 3000,

Decreasing

: 3000 < x < \infty

Plug

x = 3000

into

- 0.05 x^{2} + 300 x + 2000 : 452000

Maximum (3000, 452000)

f (x, y) = e^{x^{2}} + y^{2} + 1

f (x) = - x^{(3)} + 12 x^{(2)} + 45 x - 52

f (x) = 2 cos (x) + x

f (x) = 2 cos (x) - 1

f (x, y) = y ln (5 x + 9 y)

What is the extreme f(x)=-0.05x^2+300x+2000 ?
The extreme f(x)=-0.05x^2+300x+2000 is Maximum(3000,452000)