Algebra Examples

y = arccos (x)

$y = \arccos (x)$

Step 1

Set the argument in

arccos (x)

$\arccos (x)$ greater than or equal to

- 1

$- 1$ to find where the expression is defined.

x \geq - 1

$x \geq - 1$

Step 2

Set the argument in

arccos (x)

$\arccos (x)$ less than or equal to

1

$1$ to find where the expression is defined.

x \leq 1

$x \leq 1$

Step 3

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

[- 1, 1]

$[- 1, 1]$

Set-Builder Notation:

{x | - 1 \leq x \leq 1}

${x | - 1 \leq x \leq 1}$

Step 4

The range is the set of all valid

y

$y$ values. Use the graph to find the range.

Interval Notation:

[0, π]

$[0, π]$

Set-Builder Notation:

{y | 0 \leq y \leq π}

${y | 0 \leq y \leq π}$

Step 5

Determine the domain and range.

Domain:

[- 1, 1], {x | - 1 \leq x \leq 1}

$[- 1, 1], {x | - 1 \leq x \leq 1}$

Range:

[0, π], {y | 0 \leq y \leq π}

$[0, π], {y | 0 \leq y \leq π}$

Step 6

⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

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