What is the extreme f(x)=0.2x^2-20x+900 ?

The extreme f(x)=0.2x^2-20x+900 is Minimum(50,400)

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extreme f(x)=0.2x^2-20x+900

extreme points

f (x) = 0.2 x^{2} - 20 x + 900

Minimum (50, 400)

Solution steps

f^{'} (x) = 0.4 x - 20

Find intervals:

Decreasing

: - \infty < x < 50,

Increasing

: 50 < x < \infty

Plug

x = 50

into

0.2 x^{2} - 20 x + 900 : 400

Minimum (50, 400)

14 + 4 x - x^{2}

f (x) = - x (x + 1)

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

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

f (x) = x^{3} + 5 x^{2} - 8 x + 1

What is the extreme f(x)=0.2x^2-20x+900 ?
The extreme f(x)=0.2x^2-20x+900 is Minimum(50,400)