domain of f(x)=sqrt(x-9)+sqrt(x+14)

We save the definition:

f (x) = \sqrt{x - 9} + \sqrt{x + 14}

Let's find the domain of

\sqrt{x - 9} + \sqrt{x + 14}

For the expression

\sqrt{x - 9}

to be defined, we must have:

x - 9 \geq 0

For the expression

\sqrt{x + 14}

to be defined, we must have:

x + 14 \geq 0

Let's solve in

ℝ

x - 9 \geq 0

ⓘ We add

9

to both sides

Solution found:

x \geq 9

Let's solve in

ℝ

x + 14 \geq 0

ⓘ We subtract

14

from both sides

Solution found:

x \geq -14

Intersecting

x \geq 9

and

x \geq -14

gives

x \geq 9

▶Domain of definition:

x \geq 9