Trigonometry Examples

$\frac{2 x}{\sqrt{x - 5}}$

Step 1

Find the domain for $y = \frac{2 x}{\sqrt{x - 5}}$ 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 $\sqrt{x - 5}$ greater than or equal to $0$ to find where the expression is defined.

$x - 5 \geq 0$

Step 1.2

Add $5$ to both sides of the inequality.

$x \geq 5$

Step 1.3

Set the denominator in $\frac{2 x \sqrt{x - 5}}{x - 5}$ equal to $0$ to find where the expression is undefined.

$x - 5 = 0$

Step 1.4

Add $5$ to both sides of the equation.

$x = 5$

Step 1.5

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

Interval Notation:

$(5, \infty)$

Set-Builder Notation:

${x | x > 5}$

Interval Notation:

$(5, \infty)$

Set-Builder Notation:

${x | x > 5}$

Step 2

To find the radical expression end point, substitute the $x$ value $5$ , which is the least value in the domain, into $f (x) = \frac{2 x \sqrt{x - 5}}{x - 5}$ .

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Step 2.1

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

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

Step 2.2

Subtract $5$ from $5$ .

$f (5) = \frac{2 (5) \sqrt{(5) - 5}}{0}$

Step 2.3

The expression contains a division by $0$ . The expression is undefined.

$f (5) = Undefined$

Undefined

Step 3

The radical expression end point is $(5, Undefined)$ .

$(5, Undefined)$

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 $6$ into $f (x) = \frac{2 x \sqrt{x - 5}}{x - 5}$ . In this case, the point is $(6, 12)$ .

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Step 4.1.1

Replace the variable $x$ with $6$ in the expression.

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify the expression.

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Step 4.1.2.1.1

Multiply $2$ by $6$ .

$f (6) = \frac{12 \sqrt{6 - 5}}{6 - 5}$

Step 4.1.2.1.2

Subtract $5$ from $6$ .

$f (6) = \frac{12 \sqrt{6 - 5}}{1}$

Step 4.1.2.2

Simplify the numerator.

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Step 4.1.2.2.1

Subtract $5$ from $6$ .

$f (6) = \frac{12 \sqrt{1}}{1}$

Step 4.1.2.2.2

Any root of $1$ is $1$ .

$f (6) = \frac{12 \cdot 1}{1}$

Step 4.1.2.3

Simplify the expression.

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Step 4.1.2.3.1

Multiply $12$ by $1$ .

$f (6) = \frac{12}{1}$

Step 4.1.2.3.2

Divide $12$ by $1$ .

$f (6) = 12$

Step 4.1.2.4

The final answer is $12$ .

$y = 12$

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Simplify the expression.

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Step 4.2.2.1.1

Multiply $2$ by $7$ .

$f (7) = \frac{14 \sqrt{7 - 5}}{7 - 5}$

Step 4.2.2.1.2

Subtract $5$ from $7$ .

$f (7) = \frac{14 \sqrt{7 - 5}}{2}$

Step 4.2.2.1.3

Subtract $5$ from $7$ .

$f (7) = \frac{14 \sqrt{2}}{2}$

Step 4.2.2.2

Cancel the common factor of $14$ and $2$ .

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Step 4.2.2.2.1

Factor $2$ out of $14 \sqrt{2}$ .

$f (7) = \frac{2 (7 \sqrt{2})}{2}$

Step 4.2.2.2.2

Cancel the common factors.

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Step 4.2.2.2.2.1

Factor $2$ out of $2$ .

$f (7) = \frac{2 (7 \sqrt{2})}{2 (1)}$

Step 4.2.2.2.2.2

Cancel the common factor.

$f (7) = \frac{2 (7 \sqrt{2})}{2 \cdot 1}$

Step 4.2.2.2.2.3

Rewrite the expression.

$f (7) = \frac{7 \sqrt{2}}{1}$

Step 4.2.2.2.2.4

Divide $7 \sqrt{2}$ by $1$ .

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

Step 4.2.2.3

The final answer is $7 \sqrt{2}$ .

$y = 7 \sqrt{2}$

Step 4.3

The square root can be graphed using the points around the vertex $(5, Undefined), (6, 12), (7, 9.9)$

$\begin{array}{c} x & y \\ 5 & Undefined \\ 6 & 12 \\ 7 & 9.9 \end{array}$

Step 5