Find the Inverse y=4x^2+3

$y = 4 x^{2} + 3$

Step 1

Interchange the variables.

$x = 4 y^{2} + 3$

Step 2

Solve for $y$ .

Rewrite the equation as $4 y^{2} + 3 = x$ .

$4 y^{2} + 3 = x$

Subtract $3$ from both sides of the equation.

$4 y^{2} = x - 3$

Divide each term in $4 y^{2} = x - 3$ by $4$ and simplify.

Divide each term in $4 y^{2} = x - 3$ by $4$ .

$\frac{4 y^{2}}{4} = \frac{x}{4} + \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y^{2}}{4} = \frac{x}{4} + \frac{- 3}{4}$

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

$y^{2} = \frac{x}{4} + \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$y^{2} = \frac{x}{4} - \frac{3}{4}$

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

$y = \pm \sqrt{\frac{x}{4} - \frac{3}{4}}$

Simplify $\pm \sqrt{\frac{x}{4} - \frac{3}{4}}$ .

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{x - 3}{4}}$

Rewrite $\frac{x - 3}{4}$ as ${(\frac{1}{2})}^{2} (x - 3)$ .

Factor the perfect power $1^{2}$ out of $x - 3$ .

$y = \pm \sqrt{\frac{1^{2} (x - 3)}{4}}$

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

$y = \pm \sqrt{\frac{1^{2} (x - 3)}{2^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} (x - 3)}{2^{2} \cdot 1}$ .

$y = \pm \sqrt{{(\frac{1}{2})}^{2} (x - 3)}$

Pull terms out from under the radical.

$y = \pm \frac{1}{2} \sqrt{x - 3}$

Combine $\frac{1}{2}$ and $\sqrt{x - 3}$ .

$y = \pm \frac{\sqrt{x - 3}}{2}$

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.

$y = \frac{\sqrt{x - 3}}{2}$

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

$y = - \frac{\sqrt{x - 3}}{2}$

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

$y = \frac{\sqrt{x - 3}}{2}$

$y = - \frac{\sqrt{x - 3}}{2}$

$y = \frac{\sqrt{x - 3}}{2}$

$y = - \frac{\sqrt{x - 3}}{2}$

$y = \frac{\sqrt{x - 3}}{2}$

$y = - \frac{\sqrt{x - 3}}{2}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt{x - 3}}{2}, - \frac{\sqrt{x - 3}}{2}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt{x - 3}}{2}, - \frac{\sqrt{x - 3}}{2}$ is the inverse of $f (x) = 4 x^{2} + 3$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 4 x^{2} + 3$ and $f^{- 1} (x) = \frac{\sqrt{x - 3}}{2}, - \frac{\sqrt{x - 3}}{2}$ and compare them.

Find the range of $f (x) = 4 x^{2} + 3$ .

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

Interval Notation:

$[3, \infty)$

Find the domain of $\frac{\sqrt{x - 3}}{2}$ .

Set the radicand in $\sqrt{x - 3}$ greater than or equal to $0$ to find where the expression is defined.

$x - 3 \geq 0$

Add $3$ to both sides of the inequality.

$x \geq 3$

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

$[3, \infty)$

Find the domain of $f (x) = 4 x^{2} + 3$ .

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 $f^{- 1} (x) = \frac{\sqrt{x - 3}}{2}, - \frac{\sqrt{x - 3}}{2}$ is the range of $f (x) = 4 x^{2} + 3$ and the range of $f^{- 1} (x) = \frac{\sqrt{x - 3}}{2}, - \frac{\sqrt{x - 3}}{2}$ is the domain of $f (x) = 4 x^{2} + 3$ , then $f^{- 1} (x) = \frac{\sqrt{x - 3}}{2}, - \frac{\sqrt{x - 3}}{2}$ is the inverse of $f (x) = 4 x^{2} + 3$ .

$f^{- 1} (x) = \frac{\sqrt{x - 3}}{2}, - \frac{\sqrt{x - 3}}{2}$

Step 5

Find the Inverse y=4x^2+3

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