What is the range of sqrt(x^2+8x) ?

The range of sqrt(x^2+8x) is f(x)>= 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of sqrt(x^2+8x)

range

x^{2} + 8 x

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

x^{2} + 8 x : x \leq - 8 or x \geq 0

Extreme Points of

x^{2} + 8 x :

Minimum

(- 8, 0),

Minimum

(0, 0)

Find the range for the interval

- \infty < x \leq - 8 : 0 \leq f (x) < \infty

Find the range for the interval

0 \leq x < \infty : 0 \leq f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) \geq 0 or f (x) \geq 0

f (x) \geq 0

f (x) = ln (x) + 7

5 x - 2 y = 4, (2, - 4)

\frac{1}{2 ( 4 x + 1 )}

- 6 x^{2}

4 x - 7