Calculus Examples

$y = \sqrt{49 - x^{2}}$

Step 1

Find the domain for $y = \sqrt{49 - x^{2}}$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

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Step 1.1

Set the radicand in $\sqrt{(7 + x) (7 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(7 + x) (7 - x) \geq 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$7 + x = 0$

$7 - x = 0$

Step 1.2.2

Set $7 + x$ equal to $0$ and solve for $x$ .

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Step 1.2.2.1

Set $7 + x$ equal to $0$ .

$7 + x = 0$

Step 1.2.2.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 1.2.3

Set $7 - x$ equal to $0$ and solve for $x$ .

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Step 1.2.3.1

Set $7 - x$ equal to $0$ .

$7 - x = 0$

Step 1.2.3.2

Solve $7 - x = 0$ for $x$ .

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Step 1.2.3.2.1

Subtract $7$ from both sides of the equation.

$- x = - 7$

Step 1.2.3.2.2

Divide each term in $- x = - 7$ by $- 1$ and simplify.

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Step 1.2.3.2.2.1

Divide each term in $- x = - 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 7}{- 1}$

Step 1.2.3.2.2.2

Simplify the left side.

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Step 1.2.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 7}{- 1}$

Step 1.2.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 7}{- 1}$

Step 1.2.3.2.2.3

Simplify the right side.

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Step 1.2.3.2.2.3.1

Divide $- 7$ by $- 1$ .

$x = 7$

Step 1.2.4

The final solution is all the values that make $(7 + x) (7 - x) \geq 0$ true.

$x = - 7, 7$

Step 1.2.5

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 7$

$x > 7$

Step 1.2.6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.2.6.1

Test a value on the interval $x < - 7$ to see if it makes the inequality true.

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Step 1.2.6.1.1

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Step 1.2.6.1.2

Replace $x$ with $- 10$ in the original inequality.

$(7 - 10) (7 - (- 10)) \geq 0$

Step 1.2.6.1.3

The left side $- 51$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.6.2

Test a value on the interval $- 7 < x < 7$ to see if it makes the inequality true.

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Step 1.2.6.2.1

Choose a value on the interval $- 7 < x < 7$ and see if this value makes the original inequality true.

$x = 0$

Step 1.2.6.2.2

Replace $x$ with $0$ in the original inequality.

$(7 + 0) (7 - (0)) \geq 0$

Step 1.2.6.2.3

The left side $49$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.6.3

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 1.2.6.3.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 1.2.6.3.2

Replace $x$ with $10$ in the original inequality.

$(7 + 10) (7 - (10)) \geq 0$

Step 1.2.6.3.3

The left side $- 51$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.6.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 7$ False

$- 7 < x < 7$ True

$x > 7$ False

$x < - 7$ False

$- 7 < x < 7$ True

$x > 7$ False

Step 1.2.7

The solution consists of all of the true intervals.

$- 7 \leq x \leq 7$

Step 1.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[- 7, 7]$

Set-Builder Notation:

${x | - 7 \leq x \leq 7}$

Interval Notation:

$[- 7, 7]$

Set-Builder Notation:

${x | - 7 \leq x \leq 7}$

Step 2

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = \sqrt{(7 + x) (7 - x)}$ .

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Step 2.1

Replace the variable $x$ with $- 7$ in the expression.

$f (- 7) = \sqrt{(7 - 7) (7 - (- 7))}$

Step 2.2

Simplify the result.

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Step 2.2.1

Remove parentheses.

$f (- 7) = \sqrt{(7 - 7) (7 - (- 7))}$

Step 2.2.2

Subtract $7$ from $7$ .

$f (- 7) = \sqrt{0 (7 - (- 7))}$

Step 2.2.3

Multiply $- 1$ by $- 7$ .

$f (- 7) = \sqrt{0 (7 + 7)}$

Step 2.2.4

Add $7$ and $7$ .

$f (- 7) = \sqrt{0 \cdot 14}$

Step 2.2.5

Multiply $0$ by $14$ .

$f (- 7) = \sqrt{0}$

Step 2.2.6

Rewrite $0$ as $0^{2}$ .

$f (- 7) = \sqrt{0^{2}}$

Step 2.2.7

Pull terms out from under the radical, assuming positive real numbers.

$f (- 7) = 0$

Step 2.2.8

The final answer is $0$ .

$0$

Step 2.3

Replace the variable $x$ with $7$ in the expression.

$f (7) = \sqrt{(7 + 7) (7 - (7))}$

Step 2.4

Simplify the result.

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Step 2.4.1

Remove parentheses.

$f (7) = \sqrt{(7 + 7) (7 - (7))}$

Step 2.4.2

Add $7$ and $7$ .

$f (7) = \sqrt{14 (7 - (7))}$

Step 2.4.3

Multiply $- 1$ by $7$ .

$f (7) = \sqrt{14 (7 - 7)}$

Step 2.4.4

Subtract $7$ from $7$ .

$f (7) = \sqrt{14 \cdot 0}$

Step 2.4.5

Multiply $14$ by $0$ .

$f (7) = \sqrt{0}$

Step 2.4.6

Rewrite $0$ as $0^{2}$ .

$f (7) = \sqrt{0^{2}}$

Step 2.4.7

Pull terms out from under the radical, assuming positive real numbers.

$f (7) = 0$

Step 2.4.8

The final answer is $0$ .

$0$

Step 3

The end points are $(- 7, 0), (7, 0)$ .

$(- 7, 0), (7, 0)$

Step 4

Select a few $x$ values from the domain. It would be more useful to select the values so that they are next to the $x$ value of the radical expression end point.

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Step 4.1

Substitute the $x$ value $- 5$ into $f (x) = \sqrt{(7 + x) (7 - x)}$ . In this case, the point is $(- 5, 2 \sqrt{6})$ .

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Step 4.1.1

Replace the variable $x$ with $- 5$ in the expression.

$f (- 5) = \sqrt{(7 - 5) (7 - (- 5))}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Remove parentheses.

$f (- 5) = \sqrt{(7 - 5) (7 - (- 5))}$

Step 4.1.2.2

Subtract $5$ from $7$ .

$f (- 5) = \sqrt{2 (7 - (- 5))}$

Step 4.1.2.3

Multiply $- 1$ by $- 5$ .

$f (- 5) = \sqrt{2 (7 + 5)}$

Step 4.1.2.4

Add $7$ and $5$ .

$f (- 5) = \sqrt{2 \cdot 12}$

Step 4.1.2.5

Multiply $2$ by $12$ .

$f (- 5) = \sqrt{24}$

Step 4.1.2.6

Rewrite $24$ as $2^{2} \cdot 6$ .

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Step 4.1.2.6.1

Factor $4$ out of $24$ .

$f (- 5) = \sqrt{4 (6)}$

Step 4.1.2.6.2

Rewrite $4$ as $2^{2}$ .

$f (- 5) = \sqrt{2^{2} \cdot 6}$

Step 4.1.2.7

Pull terms out from under the radical.

$f (- 5) = 2 \sqrt{6}$

Step 4.1.2.8

The final answer is $2 \sqrt{6}$ .

$y = 2 \sqrt{6}$

Step 4.2

Substitute the $x$ value $- 3$ into $f (x) = \sqrt{(7 + x) (7 - x)}$ . In this case, the point is $(- 3, 2 \sqrt{10})$ .

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Step 4.2.1

Replace the variable $x$ with $- 3$ in the expression.

$f (- 3) = \sqrt{(7 - 3) (7 - (- 3))}$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Remove parentheses.

$f (- 3) = \sqrt{(7 - 3) (7 - (- 3))}$

Step 4.2.2.2

Subtract $3$ from $7$ .

$f (- 3) = \sqrt{4 (7 - (- 3))}$

Step 4.2.2.3

Multiply $- 1$ by $- 3$ .

$f (- 3) = \sqrt{4 (7 + 3)}$

Step 4.2.2.4

Add $7$ and $3$ .

$f (- 3) = \sqrt{4 \cdot 10}$

Step 4.2.2.5

Multiply $4$ by $10$ .

$f (- 3) = \sqrt{40}$

Step 4.2.2.6

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 4.2.2.6.1

Factor $4$ out of $40$ .

$f (- 3) = \sqrt{4 (10)}$

Step 4.2.2.6.2

Rewrite $4$ as $2^{2}$ .

$f (- 3) = \sqrt{2^{2} \cdot 10}$

Step 4.2.2.7

Pull terms out from under the radical.

$f (- 3) = 2 \sqrt{10}$

Step 4.2.2.8

The final answer is $2 \sqrt{10}$ .

$y = 2 \sqrt{10}$

Step 4.3

Substitute the $x$ value $- 1$ into $f (x) = \sqrt{(7 + x) (7 - x)}$ . In this case, the point is $(- 1, 4 \sqrt{3})$ .

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Step 4.3.1

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

$f (- 1) = \sqrt{(7 - 1) (7 - (- 1))}$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Remove parentheses.

$f (- 1) = \sqrt{(7 - 1) (7 - (- 1))}$

Step 4.3.2.2

Subtract $1$ from $7$ .

$f (- 1) = \sqrt{6 (7 - (- 1))}$

Step 4.3.2.3

Multiply $- 1$ by $- 1$ .

$f (- 1) = \sqrt{6 (7 + 1)}$

Step 4.3.2.4

Add $7$ and $1$ .

$f (- 1) = \sqrt{6 \cdot 8}$

Step 4.3.2.5

Multiply $6$ by $8$ .

$f (- 1) = \sqrt{48}$

Step 4.3.2.6

Rewrite $48$ as $4^{2} \cdot 3$ .

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Step 4.3.2.6.1

Factor $16$ out of $48$ .

$f (- 1) = \sqrt{16 (3)}$

Step 4.3.2.6.2

Rewrite $16$ as $4^{2}$ .

$f (- 1) = \sqrt{4^{2} \cdot 3}$

Step 4.3.2.7

Pull terms out from under the radical.

$f (- 1) = 4 \sqrt{3}$

Step 4.3.2.8

The final answer is $4 \sqrt{3}$ .

$y = 4 \sqrt{3}$

Step 4.4

Substitute the $x$ value $3$ into $f (x) = \sqrt{(7 + x) (7 - x)}$ . In this case, the point is $(3, 2 \sqrt{10})$ .

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Step 4.4.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = \sqrt{(7 + 3) (7 - (3))}$

Step 4.4.2

Simplify the result.

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Step 4.4.2.1

Remove parentheses.

$f (3) = \sqrt{(7 + 3) (7 - (3))}$

Step 4.4.2.2

Add $7$ and $3$ .

$f (3) = \sqrt{10 (7 - (3))}$

Step 4.4.2.3

Multiply $- 1$ by $3$ .

$f (3) = \sqrt{10 (7 - 3)}$

Step 4.4.2.4

Subtract $3$ from $7$ .

$f (3) = \sqrt{10 \cdot 4}$

Step 4.4.2.5

Multiply $10$ by $4$ .

$f (3) = \sqrt{40}$

Step 4.4.2.6

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 4.4.2.6.1

Factor $4$ out of $40$ .

$f (3) = \sqrt{4 (10)}$

Step 4.4.2.6.2

Rewrite $4$ as $2^{2}$ .

$f (3) = \sqrt{2^{2} \cdot 10}$

Step 4.4.2.7

Pull terms out from under the radical.

$f (3) = 2 \sqrt{10}$

Step 4.4.2.8

The final answer is $2 \sqrt{10}$ .

$y = 2 \sqrt{10}$

Step 4.5

Substitute the $x$ value $5$ into $f (x) = \sqrt{(7 + x) (7 - x)}$ . In this case, the point is $(5, 2 \sqrt{6})$ .

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Step 4.5.1

Replace the variable $x$ with $5$ in the expression.

$f (5) = \sqrt{(7 + 5) (7 - (5))}$

Step 4.5.2

Simplify the result.

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Step 4.5.2.1

Remove parentheses.

$f (5) = \sqrt{(7 + 5) (7 - (5))}$

Step 4.5.2.2

Add $7$ and $5$ .

$f (5) = \sqrt{12 (7 - (5))}$

Step 4.5.2.3

Multiply $- 1$ by $5$ .

$f (5) = \sqrt{12 (7 - 5)}$

Step 4.5.2.4

Subtract $5$ from $7$ .

$f (5) = \sqrt{12 \cdot 2}$

Step 4.5.2.5

Multiply $12$ by $2$ .

$f (5) = \sqrt{24}$

Step 4.5.2.6

Rewrite $24$ as $2^{2} \cdot 6$ .

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Step 4.5.2.6.1

Factor $4$ out of $24$ .

$f (5) = \sqrt{4 (6)}$

Step 4.5.2.6.2

Rewrite $4$ as $2^{2}$ .

$f (5) = \sqrt{2^{2} \cdot 6}$

Step 4.5.2.7

Pull terms out from under the radical.

$f (5) = 2 \sqrt{6}$

Step 4.5.2.8

The final answer is $2 \sqrt{6}$ .

$y = 2 \sqrt{6}$

Step 4.6

The square root can be graphed using the points around the vertex $(- 7, 0), (7, 0), (- 5, 4.9), (- 3, 6.32), (- 1, 6.93), (1, 6.93), (3, 6.32), (5, 4.9)$

$\begin{array}{c} x & y \\ - 7 & 0 \\ - 5 & 4.9 \\ - 3 & 6.32 \\ - 1 & 6.93 \\ 1 & 6.93 \\ 3 & 6.32 \\ 5 & 4.9 \\ 7 & 0 \end{array}$

Step 5