What is the domain of f(x)=(sqrt(2x+5))/(x^2+4x+3) ?

The domain of f(x)=(sqrt(2x+5))/(x^2+4x+3) is -5/2 -1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

domain of f(x)=(sqrt(2x+5))/(x^2+4x+3)

domain

f (x) = \frac{2 x + 5}{x ^{2} + 4 x + 3}

- \frac{5}{2} \leq x < - 1 or x > - 1

Interval Notation

[- \frac{5}{2}, - 1) \cup (- 1, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - \frac{5}{2}

Find undefined (singularity) points:

x = - 1, x = - 3

Combine real regions and undefined points for final function domain

- \frac{5}{2} \leq x < - 1 or x > - 1

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

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

g (x) = \frac{1}{x}

f (x) = lo g_{3} (2 x - 3)

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

What is the domain of f(x)=(sqrt(2x+5))/(x^2+4x+3) ?
The domain of f(x)=(sqrt(2x+5))/(x^2+4x+3) is -5/2 <= x<-1\lor x>-1