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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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)

\frac{1}{x + 4}

\frac{x ^{2} - 9}{x - 5}

\frac{x ^{3}}{( x - 2 ) ^{2}}

f (x) = 5 cos (x), 0 \leq x \leq 2 π

f (x) = \frac{1}{( x + 2 )}

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