Calculus Examples

$f (x) = | 9 - x^{2} |$

Step 1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 2

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

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Step 2.1

Substitute the $x$ value $- 2$ into $f (x) = | 9 - x^{2} |$ . In this case, the point is $(- 2, 5)$ .

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Step 2.1.1

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = | 9 - {(- 2)}^{2} |$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

Raise $- 2$ to the power of $2$ .

$f (- 2) = | 9 - 1 \cdot 4 |$

Step 2.1.2.1.2

Multiply $- 1$ by $4$ .

$f (- 2) = | 9 - 4 |$

Step 2.1.2.2

Subtract $4$ from $9$ .

$f (- 2) = | 5 |$

Step 2.1.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$f (- 2) = 5$

Step 2.1.2.4

The final answer is $5$ .

$y = 5$

Step 2.2

Substitute the $x$ value $- 1$ into $f (x) = | 9 - x^{2} |$ . In this case, the point is $(- 1, 8)$ .

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Step 2.2.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = | 9 - {(- 1)}^{2} |$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

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

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Step 2.2.2.1.1.1

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

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Step 2.2.2.1.1.1.1

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

$f (- 1) = | 9 + (- 1) {(- 1)}^{2} |$

Step 2.2.2.1.1.1.2

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

$f (- 1) = | 9 + {(- 1)}^{1 + 2} |$

Step 2.2.2.1.1.2

Add $1$ and $2$ .

$f (- 1) = | 9 + {(- 1)}^{3} |$

Step 2.2.2.1.2

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

$f (- 1) = | 9 - 1 |$

Step 2.2.2.2

Subtract $1$ from $9$ .

$f (- 1) = | 8 |$

Step 2.2.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $8$ is $8$ .

$f (- 1) = 8$

Step 2.2.2.4

The final answer is $8$ .

$y = 8$

Step 2.3

Substitute the $x$ value $0$ into $f (x) = | 9 - x^{2} |$ . In this case, the point is $(0, 9)$ .

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Step 2.3.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = | 9 - {(0)}^{2} |$

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Simplify each term.

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Step 2.3.2.1.1

Raising $0$ to any positive power yields $0$ .

$f (0) = | 9 - 0 |$

Step 2.3.2.1.2

Multiply $- 1$ by $0$ .

$f (0) = | 9 + 0 |$

Step 2.3.2.2

Add $9$ and $0$ .

$f (0) = | 9 |$

Step 2.3.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $9$ is $9$ .

$f (0) = 9$

Step 2.3.2.4

The final answer is $9$ .

$y = 9$

Step 2.4

Substitute the $x$ value $1$ into $f (x) = | 9 - x^{2} |$ . In this case, the point is $(1, 8)$ .

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Step 2.4.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = | 9 - {(1)}^{2} |$

Step 2.4.2

Simplify the result.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

One to any power is one.

$f (1) = | 9 - 1 \cdot 1 |$

Step 2.4.2.1.2

Multiply $- 1$ by $1$ .

$f (1) = | 9 - 1 |$

Step 2.4.2.2

Subtract $1$ from $9$ .

$f (1) = | 8 |$

Step 2.4.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $8$ is $8$ .

$f (1) = 8$

Step 2.4.2.4

The final answer is $8$ .

$y = 8$

Step 2.5

The absolute value can be graphed using the points around the vertex $(- 2, 5), (- 1, 8), (0, 9), (1, 8)$

$\begin{array}{c} x & y \\ - 2 & 5 \\ - 1 & 8 \\ 0 & 9 \\ 1 & 8 \end{array}$

Step 3