What is the extreme f(x)=0.75x^2-30x+800 ?

The extreme f(x)=0.75x^2-30x+800 is Minimum(20,500)

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extreme f(x)=0.75x^2-30x+800

extreme points

f (x) = 0.75 x^{2} - 30 x + 800

Minimum (20, 500)

Solution steps

f^{'} (x) = 1.5 x - 30

Find intervals:

Decreasing

: - \infty < x < 20,

Increasing

: 20 < x < \infty

Plug

x = 20

into

0.75 x^{2} - 30 x + 800 : 500

Minimum (20, 500)

What is the extreme f(x)=0.75x^2-30x+800 ?
The extreme f(x)=0.75x^2-30x+800 is Minimum(20,500)