What is the domain of f(x)=sqrt(x^2+7x-12) ?

The domain of f(x)=sqrt(x^2+7x-12) is x = (sqrt(97)-7)/2

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^2+7x-12)

domain

f (x) = x^{2} + 7 x - 12

x \leq \frac{- 97 - 7}{2} or x \geq \frac{97 - 7}{2}

Interval Notation

(- \infty, \frac{- 97 - 7}{2}] \cup [\frac{97 - 7}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq \frac{- 97 - 7}{2} or x \geq \frac{97 - 7}{2}

The function domain

x \leq \frac{- 97 - 7}{2} or x \geq \frac{97 - 7}{2}

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

f (x) = 2 x + 3 x

y = \frac{x ^{2} - 6}{2}

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

\frac{1 - x}{x + 2}

What is the domain of f(x)=sqrt(x^2+7x-12) ?
The domain of f(x)=sqrt(x^2+7x-12) is x<= (-sqrt(97)-7)/2 \lor x>= (sqrt(97)-7)/2