What is the extreme f(x)=-x^2+120x-3200 ?

The extreme f(x)=-x^2+120x-3200 is Maximum(60,400)

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extreme f(x)=-x^2+120x-3200

extreme points

f (x) = - x^{2} + 120 x - 3200

Maximum (60, 400)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 60,

Decreasing

: 60 < x < \infty

Plug

x = 60

into

- x^{2} + 120 x - 3200 : 400

Maximum (60, 400)

2 x^{3} - 3 x^{2} - 36 x + 54

2 x^{2} - x y + y^{2} + 7 x

f (x, y) = x^{3} + 2 x y^{2} + \frac{y ^{4}}{x}

f (x) = x^{4} - 242 x^{2} - 5

f (x) = 20 x - 10 x - 2

What is the extreme f(x)=-x^2+120x-3200 ?
The extreme f(x)=-x^2+120x-3200 is Maximum(60,400)