Precalculus Examples

$f (x) = - 4 x^{2} + 1$

Step 1

Write $f (x) = - 4 x^{2} + 1$ as an equation.

$y = - 4 x^{2} + 1$

Step 2

Interchange the variables.

$x = - 4 y^{2} + 1$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $- 4 y^{2} + 1 = x$ .

$- 4 y^{2} + 1 = x$

Step 3.2

Subtract $1$ from both sides of the equation.

$- 4 y^{2} = x - 1$

Step 3.3

Divide each term in $- 4 y^{2} = x - 1$ by $- 4$ and simplify.

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Step 3.3.1

Divide each term in $- 4 y^{2} = x - 1$ by $- 4$ .

$\frac{- 4 y^{2}}{- 4} = \frac{x}{- 4} + \frac{- 1}{- 4}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $- 4$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{- 4 y^{2}}{- 4} = \frac{x}{- 4} + \frac{- 1}{- 4}$

Step 3.3.2.1.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{x}{- 4} + \frac{- 1}{- 4}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Move the negative in front of the fraction.

$y^{2} = - \frac{x}{4} + \frac{- 1}{- 4}$

Step 3.3.3.1.2

Dividing two negative values results in a positive value.

$y^{2} = - \frac{x}{4} + \frac{1}{4}$

Step 3.4

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

$y = \pm \sqrt{- \frac{x}{4} + \frac{1}{4}}$

Step 3.5

Simplify $\pm \sqrt{- \frac{x}{4} + \frac{1}{4}}$ .

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Step 3.5.1

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{- x + 1}{4}}$

Step 3.5.2

Rewrite $\frac{- x + 1}{4}$ as ${(\frac{1}{2})}^{2} (- x + 1)$ .

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Step 3.5.2.1

Factor the perfect power $1^{2}$ out of $- x + 1$ .

$y = \pm \sqrt{\frac{1^{2} (- x + 1)}{4}}$

Step 3.5.2.2

Factor the perfect power $2^{2}$ out of $4$ .

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

Step 3.5.2.3

Rearrange the fraction $\frac{1^{2} (- x + 1)}{2^{2} \cdot 1}$ .

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

Step 3.5.3

Pull terms out from under the radical.

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

Step 3.5.4

Combine $\frac{1}{2}$ and $\sqrt{- x + 1}$ .

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

Step 3.6

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

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Step 3.6.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{\sqrt{- x + 1}}{2}$

Step 3.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \frac{\sqrt{- x + 1}}{2}$

Step 3.6.3

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

$y = \frac{\sqrt{- x + 1}}{2}$

$y = - \frac{\sqrt{- x + 1}}{2}$

$y = \frac{\sqrt{- x + 1}}{2}$

$y = - \frac{\sqrt{- x + 1}}{2}$

$y = \frac{\sqrt{- x + 1}}{2}$

$y = - \frac{\sqrt{- x + 1}}{2}$

Step 4

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

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

Step 5

Verify if $f^{- 1} (x) = \frac{\sqrt{- x + 1}}{2}, - \frac{\sqrt{- x + 1}}{2}$ is the inverse of $f (x) = - 4 x^{2} + 1$ .

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Step 5.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = - 4 x^{2} + 1$ and $f^{- 1} (x) = \frac{\sqrt{- x + 1}}{2}, - \frac{\sqrt{- x + 1}}{2}$ and compare them.

Step 5.2

Find the range of $f (x) = - 4 x^{2} + 1$ .

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Step 5.2.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, 1]$

Step 5.3

Find the domain of $\frac{\sqrt{- x + 1}}{2}$ .

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Step 5.3.1

Set the radicand in $\sqrt{- x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$- x + 1 \geq 0$

Step 5.3.2

Solve for $x$ .

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Step 5.3.2.1

Subtract $1$ from both sides of the inequality.

$- x \geq - 1$

Step 5.3.2.2

Divide each term in $- x \geq - 1$ by $- 1$ and simplify.

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Step 5.3.2.2.1

Divide each term in $- x \geq - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 1}{- 1}$

Step 5.3.2.2.2

Simplify the left side.

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Step 5.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 1}{- 1}$

Step 5.3.2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 1}{- 1}$

Step 5.3.2.2.3

Simplify the right side.

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Step 5.3.2.2.3.1

Divide $- 1$ by $- 1$ .

$x \leq 1$

Step 5.3.3

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

$(- \infty, 1]$

Step 5.4

Find the domain of $f (x) = - 4 x^{2} + 1$ .

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Step 5.4.1

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 5.5

Since the domain of $f^{- 1} (x) = \frac{\sqrt{- x + 1}}{2}, - \frac{\sqrt{- x + 1}}{2}$ is the range of $f (x) = - 4 x^{2} + 1$ and the range of $f^{- 1} (x) = \frac{\sqrt{- x + 1}}{2}, - \frac{\sqrt{- x + 1}}{2}$ is the domain of $f (x) = - 4 x^{2} + 1$ , then $f^{- 1} (x) = \frac{\sqrt{- x + 1}}{2}, - \frac{\sqrt{- x + 1}}{2}$ is the inverse of $f (x) = - 4 x^{2} + 1$ .

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

Step 6