What is the extreme f(x)=-0.04x^2+42x-500 ?

The extreme f(x)=-0.04x^2+42x-500 is Maximum(525,10525)

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extreme f(x)=-0.04x^2+42x-500

extreme points

f (x) = - 0.04 x^{2} + 42 x - 500

Maximum (525, 10525)

Solution steps

f^{'} (x) = - 0.08 x + 42

Find intervals:

Increasing

: - \infty < x < 525,

Decreasing

: 525 < x < \infty

Plug

x = 525

into

- 0.04 x^{2} + 42 x - 500 : 10525

Maximum (525, 10525)

f (x) = x^{5} - 10 x^{4} + 2 = x^{4} (x - 10) + 2

f (x) = x^{9} - x^{7}

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

f (x) = ln (4 x) - 8 x^{2}

3 x^{4} + 3 y^{4} - 2 x y

What is the extreme f(x)=-0.04x^2+42x-500 ?
The extreme f(x)=-0.04x^2+42x-500 is Maximum(525,10525)