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Algebra Examples

f (x) = \sqrt{x}

$f (x) = \sqrt{x}$ ,

g (x) = x + 1

$g (x) = x + 1$ ,

f (g (x))

$f (g (x))$

Step 1

Set up the composite result function.

f (g (x))

$f (g (x))$

Step 2

Evaluate

f (x + 1)

$f (x + 1)$ by substituting in the value of

g

$g$ into

f

$f$ .

f (x + 1) = \sqrt{x + 1}

$f (x + 1) = \sqrt{x + 1}$

Step 3

Remove parentheses.

\sqrt{x + 1}

$\sqrt{x + 1}$

Step 4

Set the radicand in

\sqrt{x + 1}

$\sqrt{x + 1}$ greater than or equal to

0

$0$ to find where the expression is defined.

x + 1 \geq 0

$x + 1 \geq 0$

Step 5

Subtract

1

$1$ from both sides of the inequality.

x \geq - 1

$x \geq - 1$

Step 6

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

[- 1, \infty)

$[- 1, \infty)$

Set-Builder Notation:

{x | x \geq - 1}

${x | x \geq - 1}$

Step 7

Enter YOUR Problem

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$