Precalculus Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

Step 2.3.2.1

Cancel the common factor of $2$ .

Tap for more steps...

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

Tap for more steps...

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.

Tap for more steps...

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 $2$ .

Tap for more steps...

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 $2$ .

Tap for more steps...

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.

Tap for more steps...

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

Tap for more steps...

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

Tap for more steps...

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

Tap for more steps...

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 $x$ .

Tap for more steps...

Step 4.3.2.1

Divide each term in $2 (x + 5) \geq 0$ by $2$ and simplify.

Tap for more steps...

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.

Tap for more steps...

Step 4.3.2.1.2.1

Cancel the common factor of $2$ .

Tap for more steps...

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.

Tap for more steps...

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

Tap for more steps...

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