Precalculus Examples

$y = 2 x^{2} - 5$

Step 1

Interchange the variables.

$x = 2 y^{2} - 5$

Step 2

Solve for

$y$ .

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Step 2.1

Rewrite the equation as

$2 y^{2} - 5 = x$ .

$2 y^{2} - 5 = x$

Step 2.2

Add

$5$ to both sides of the equation.

$2 y^{2} = x + 5$

Step 2.3

Divide each term in

$2 y^{2} = x + 5$ by

$2$ and simplify.

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Step 2.3.1

Divide each term in

$2 y^{2} = x + 5$ by

$2$ .

$\frac{2 y^{2}}{2} = \frac{x}{2} + \frac{5}{2}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of

$2$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{x}{2} + \frac{5}{2}$

Step 2.3.2.1.2

Divide

$y^{2}$ by

$1$ .

$y^{2} = \frac{x}{2} + \frac{5}{2}$

Step 2.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\frac{x}{2} + \frac{5}{2}}$

Step 2.5

Simplify

$\pm \sqrt{\frac{x}{2} + \frac{5}{2}}$ .

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Step 2.5.1

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{x + 5}{2}}$

Step 2.5.2

Rewrite

$\sqrt{\frac{x + 5}{2}}$ as

$\frac{\sqrt{x + 5}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x + 5}}{\sqrt{2}}$

Step 2.5.3

Multiply

$\frac{\sqrt{x + 5}}{\sqrt{2}}$ by

$\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x + 5}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Step 2.5.4

Combine and simplify the denominator.

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Step 2.5.4.1

Multiply

$\frac{\sqrt{x + 5}}{\sqrt{2}}$ by

$\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

Step 2.5.4.2

Raise

$\sqrt{2}$ to the power of

$1$ .

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Step 2.5.4.3

Raise

$\sqrt{2}$ to the power of

$1$ .

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Step 2.5.4.4

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Step 2.5.4.5

Add

$1$ and

$1$ .

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{{\sqrt{2}}^{2}}$

Step 2.5.4.6

Rewrite

${\sqrt{2}}^{2}$ as

$2$ .

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Step 2.5.4.6.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{2}$ as

$2^{\frac{1}{2}}$ .

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Step 2.5.4.6.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Step 2.5.4.6.3

Combine

$\frac{1}{2}$ and

$2$ .

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{2^{\frac{2}{2}}}$

Step 2.5.4.6.4

Cancel the common factor of

$2$ .

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Step 2.5.4.6.4.1

Cancel the common factor.

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{2^{\frac{2}{2}}}$

Step 2.5.4.6.4.2

Rewrite the expression.

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{2^{1}}$

Step 2.5.4.6.5

Evaluate the exponent.

$y = \pm \frac{\sqrt{x + 5} \sqrt{2}}{2}$

Step 2.5.5

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{(x + 5) \cdot 2}}{2}$

Step 2.5.6

Reorder factors in

$\pm \frac{\sqrt{(x + 5) \cdot 2}}{2}$ .

$y = \pm \frac{\sqrt{2 (x + 5)}}{2}$

Step 2.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.6.1

First, use the positive value of the

$\pm$ to find the first solution.

$y = \frac{\sqrt{2 (x + 5)}}{2}$

Step 2.6.2

Next, use the negative value of the

$\pm$ to find the second solution.

$y = - \frac{\sqrt{2 (x + 5)}}{2}$

Step 2.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{\sqrt{2 (x + 5)}}{2}$

$y = - \frac{\sqrt{2 (x + 5)}}{2}$

$y = \frac{\sqrt{2 (x + 5)}}{2}$

$y = - \frac{\sqrt{2 (x + 5)}}{2}$

$y = \frac{\sqrt{2 (x + 5)}}{2}$

$y = - \frac{\sqrt{2 (x + 5)}}{2}$

Step 3

Replace

$y$ with

$f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{\sqrt{2 (x + 5)}}{2}, - \frac{\sqrt{2 (x + 5)}}{2}$

Step 4

Verify if

$f^{- 1} (x) = \frac{\sqrt{2 (x + 5)}}{2}, - \frac{\sqrt{2 (x + 5)}}{2}$ is the inverse of

$f (x) = 2 x^{2} - 5$ .

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Step 4.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of

$f (x) = 2 x^{2} - 5$ and

$f^{- 1} (x) = \frac{\sqrt{2 (x + 5)}}{2}, - \frac{\sqrt{2 (x + 5)}}{2}$ and compare them.

Step 4.2

Find the range of

$f (x) = 2 x^{2} - 5$ .

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Step 4.2.1

The range is the set of all valid

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

Interval Notation:

$[- 5, \infty)$

Step 4.3

Find the domain of

$\frac{\sqrt{2 (x + 5)}}{2}$ .

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Step 4.3.1

Set the radicand in

$\sqrt{2 (x + 5)}$ greater than or equal to

$0$ to find where the expression is defined.

$2 (x + 5) \geq 0$

Step 4.3.2

Solve for

$x$ .

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Step 4.3.2.1

Divide each term in

$2 (x + 5) \geq 0$ by

$2$ and simplify.

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Step 4.3.2.1.1

Divide each term in

$2 (x + 5) \geq 0$ by

$2$ .

$\frac{2 (x + 5)}{2} \geq \frac{0}{2}$

Step 4.3.2.1.2

Simplify the left side.

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Step 4.3.2.1.2.1

Cancel the common factor of

$2$ .

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Step 4.3.2.1.2.1.1

Cancel the common factor.

$\frac{2 (x + 5)}{2} \geq \frac{0}{2}$

Step 4.3.2.1.2.1.2

Divide

$x + 5$ by

$1$ .

$x + 5 \geq \frac{0}{2}$

Step 4.3.2.1.3

Simplify the right side.

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Step 4.3.2.1.3.1

Divide

$0$ by

$2$ .

$x + 5 \geq 0$

Step 4.3.2.2

Subtract

$5$ from both sides of the inequality.

$x \geq - 5$

Step 4.3.3

The domain is all values of

$x$ that make the expression defined.

$[- 5, \infty)$

Step 4.4

Find the domain of

$f (x) = 2 x^{2} - 5$ .

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Step 4.4.1

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.

$(- \infty, \infty)$

Step 4.5

Since the domain of

$f^{- 1} (x) = \frac{\sqrt{2 (x + 5)}}{2}, - \frac{\sqrt{2 (x + 5)}}{2}$ is the range of

$f (x) = 2 x^{2} - 5$ and the range of

$f^{- 1} (x) = \frac{\sqrt{2 (x + 5)}}{2}, - \frac{\sqrt{2 (x + 5)}}{2}$ is the domain of

$f (x) = 2 x^{2} - 5$ , then

$f^{- 1} (x) = \frac{\sqrt{2 (x + 5)}}{2}, - \frac{\sqrt{2 (x + 5)}}{2}$ is the inverse of

$f (x) = 2 x^{2} - 5$ .

$f^{- 1} (x) = \frac{\sqrt{2 (x + 5)}}{2}, - \frac{\sqrt{2 (x + 5)}}{2}$

Step 5