What is the extreme f(x)=-1.25x^2+160x-2500 ?

The extreme f(x)=-1.25x^2+160x-2500 is Maximum(64,2620)

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extreme f(x)=-1.25x^2+160x-2500

extreme points

f (x) = - 1.25 x^{2} + 160 x - 2500

Maximum (64, 2620)

Solution steps

f^{'} (x) = - 2.5 x + 160

Find intervals:

Increasing

: - \infty < x < 64,

Decreasing

: 64 < x < \infty

Plug

x = 64

into

- 1.25 x^{2} + 160 x - 2500 : 2620

Maximum (64, 2620)

What is the extreme f(x)=-1.25x^2+160x-2500 ?
The extreme f(x)=-1.25x^2+160x-2500 is Maximum(64,2620)