What is the extreme f(x)=-2500x^2+45000+1360000 ?

The extreme f(x)=-2500x^2+45000+1360000 is Maximum(0,1405000)

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extreme f(x)=-2500x^2+45000+1360000

extreme points

f (x) = - 2500 x^{2} + 45000 + 1360000

Maximum (0, 1405000)

Solution steps

f^{'} (x) = - 5000 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = 0

into

- 2500 x^{2} + 45000 + 1360000 : 1405000

Maximum (0, 1405000)

f (x) = x + 12 x^{2} - 72 x + 3

f (x, y) = 8 x^{2} - x y + 17 x - 17 y - 18

- x^{2} - y^{2} + x + 2 y - 1

(x - 1) ln (x - 1)

lo g_{10} (1 - x^{2})

What is the extreme f(x)=-2500x^2+45000+1360000 ?
The extreme f(x)=-2500x^2+45000+1360000 is Maximum(0,1405000)