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)

f (x) = 2 x^{3} - 150 x + 3

f (x, y) = x^{2} + \frac{1}{2} y^{2} + \frac{1}{2} (y - x) - \frac{3}{2}

x^{4} + 3 x^{3} + 2 x^{2} + x + 1

f (x) = - 0.4 x^{2} + 90 x - 2000, 0 \leq x

2 x^{2} - 2 x

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