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

The domain of f(x)=(sqrt(x)+3)/(x^2+4x-5) is 0 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(x)+3)/(x^2+4x-5)

domain

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

0 \leq x < 1 or x > 1

Interval Notation

[0, 1) \cup (1, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = 1, x = - 5

Combine real regions and undefined points for final function domain

0 \leq x < 1 or x > 1

f (x) = \frac{x - 5}{x - 4 x}

f (x) = - x^{2} - 4 x - 5

f (x) = \frac{x ^{2}}{x - 1} + ln (4 - ∣ 2 x + 1 ∣)

f (x) = - x^{2} - 4 x - 6

f (x) = \frac{1}{\frac{4}{x - 3} - 2}

What is the domain of f(x)=(sqrt(x)+3)/(x^2+4x-5) ?
The domain of f(x)=(sqrt(x)+3)/(x^2+4x-5) is 0<= x<1\lor x>1