Precalculus Examples

$f (x) = {\begin{cases} - {(x + 1)}^{2} - 3 & x \leq 0 \\ - 6 & x \geq 4 \end{cases}$ Find $f (- 2)$

Step 1

Identify the piece that describes the function at $x = - 2$ . In this case, $x = - 2$ falls within the interval $x \leq 0$ , therefore use $- {(x + 1)}^{2} - 3$ to evaluate $f (- 2)$ .

$f (x) = {\begin{cases} - {(x + 1)}^{2} - 3 & x \leq 0 \\ - 6 & x \geq 4 \end{cases}$

Step 2

The function is equal to $- {(x + 1)}^{2} - 3$ at $x = - 2$ .

$f (x) = - {(x + 1)}^{2} - 3 (]_{x = - 2}^{})$

Step 3

Evaluate the function at $x = - 2$ .

$f (- 2) = - {((- 2) + 1)}^{2} - 3$

Step 4

Simplify each term.

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Step 4.1

Add $- 2$ and $1$ .

$f (- 2) = - {(- 1)}^{2} - 3$

Step 4.2

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 4.2.1

Multiply $- 1$ by ${(- 1)}^{2}$ .

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Step 4.2.1.1

Raise $- 1$ to the power of $1$ .

$f (- 2) = (- 1) {(- 1)}^{2} - 3$

Step 4.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (- 2) = {(- 1)}^{1 + 2} - 3$

Step 4.2.2

Add $1$ and $2$ .

$f (- 2) = {(- 1)}^{3} - 3$

Step 4.3

Raise $- 1$ to the power of $3$ .

$f (- 2) = - 1 - 3$

Step 5

Subtract $3$ from $- 1$ .

$f (- 2) = - 4$