What is the extreme f(x)=-0.01x^2+120x+200000 ?

The extreme f(x)=-0.01x^2+120x+200000 is Maximum(6000,560000)

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extreme f(x)=-0.01x^2+120x+200000

extreme points

f (x) = - 0.01 x^{2} + 120 x + 200000

Maximum (6000, 560000)

Solution steps

f^{'} (x) = - 0.02 x + 120

Find intervals:

Increasing

: - \infty < x < 6000,

Decreasing

: 6000 < x < \infty

Plug

x = 6000

into

- 0.01 x^{2} + 120 x + 200000 : 560000

Maximum (6000, 560000)

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

\frac{x}{ln ( x ^{2} )}

x^{2 x}

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

f (x, y) = \frac{500}{( 4 + x ^{2} + y ^{2} )}

What is the extreme f(x)=-0.01x^2+120x+200000 ?
The extreme f(x)=-0.01x^2+120x+200000 is Maximum(6000,560000)