Find the Inverse y=16x^2+1

$y = 16 x^{2} + 1$

Step 1

Interchange the variables.

$x = 16 y^{2} + 1$

Step 2

Solve for $y$ .

Rewrite the equation as $16 y^{2} + 1 = x$ .

$16 y^{2} + 1 = x$

Subtract $1$ from both sides of the equation.

$16 y^{2} = x - 1$

Divide each term in $16 y^{2} = x - 1$ by $16$ and simplify.

Divide each term in $16 y^{2} = x - 1$ by $16$ .

$\frac{16 y^{2}}{16} = \frac{x}{16} + \frac{- 1}{16}$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 y^{2}}{16} = \frac{x}{16} + \frac{- 1}{16}$

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

$y^{2} = \frac{x}{16} + \frac{- 1}{16}$

Simplify the right side.

Move the negative in front of the fraction.

$y^{2} = \frac{x}{16} - \frac{1}{16}$

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

$y = \pm \sqrt{\frac{x}{16} - \frac{1}{16}}$

Simplify $\pm \sqrt{\frac{x}{16} - \frac{1}{16}}$ .

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{x - 1}{16}}$

Rewrite $\frac{x - 1}{16}$ as ${(\frac{1}{4})}^{2} (x - 1)$ .

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

$y = \pm \sqrt{\frac{1^{2} (x - 1)}{16}}$

Factor the perfect power $4^{2}$ out of $16$ .

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

Rearrange the fraction $\frac{1^{2} (x - 1)}{4^{2} \cdot 1}$ .

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

Pull terms out from under the radical.

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

Combine $\frac{1}{4}$ and $\sqrt{x - 1}$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

Step 3

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

$f^{- 1} (x) = \frac{\sqrt{x - 1}}{4}, - \frac{\sqrt{x - 1}}{4}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt{x - 1}}{4}, - \frac{\sqrt{x - 1}}{4}$ is the inverse of $f (x) = 16 x^{2} + 1$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 16 x^{2} + 1$ and $f^{- 1} (x) = \frac{\sqrt{x - 1}}{4}, - \frac{\sqrt{x - 1}}{4}$ and compare them.

Find the range of $f (x) = 16 x^{2} + 1$ .

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

Interval Notation:

$[1, \infty)$

Find the domain of $\frac{\sqrt{x - 1}}{4}$ .

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

$x - 1 \geq 0$

Add $1$ to both sides of the inequality.

$x \geq 1$

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

$[1, \infty)$

Find the domain of $f (x) = 16 x^{2} + 1$ .

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.

$(- \infty, \infty)$

Since the domain of $f^{- 1} (x) = \frac{\sqrt{x - 1}}{4}, - \frac{\sqrt{x - 1}}{4}$ is the range of $f (x) = 16 x^{2} + 1$ and the range of $f^{- 1} (x) = \frac{\sqrt{x - 1}}{4}, - \frac{\sqrt{x - 1}}{4}$ is the domain of $f (x) = 16 x^{2} + 1$ , then $f^{- 1} (x) = \frac{\sqrt{x - 1}}{4}, - \frac{\sqrt{x - 1}}{4}$ is the inverse of $f (x) = 16 x^{2} + 1$ .

$f^{- 1} (x) = \frac{\sqrt{x - 1}}{4}, - \frac{\sqrt{x - 1}}{4}$

Step 5

Find the Inverse y=16x^2+1

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