What is the range of sqrt(4x^2+20) ?

The range of sqrt(4x^2+20) is f(x)>= 2sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of sqrt(4x^2+20)

range

4 x^{2} + 20

f (x) \geq 25

Interval Notation

[25, \infty)

Solution steps

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

Domain of

4 x^{2} + 20 : - \infty < x < \infty

Simplify

4 x^{2} + 20 : 2 x^{2} + 5

Extreme Points of

2 x^{2} + 5 :

Minimum

(0, 25)

Find the range for the interval

- \infty < x < \infty : 25 \leq f (x) < \infty

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

f (x) \geq 25

(7, - 6), (- 5, 8)

f (x) = x^{2} (7 - x)^{3}

sec^{2} (θ) d θ

f (x) = 3 \cdot e^{2 x}

f (x) = \frac{2 x - 3}{x + 1}

What is the range of sqrt(4x^2+20) ?
The range of sqrt(4x^2+20) is f(x)>= 2sqrt(5)