Find the Inverse y=1-3x^2

$y = 1 - 3 x^{2}$

Step 1

Interchange the variables.

$x = 1 - 3 y^{2}$

Step 2

Solve for $y$ .

Rewrite the equation as $1 - 3 y^{2} = x$ .

$1 - 3 y^{2} = x$

Subtract $1$ from both sides of the equation.

$- 3 y^{2} = x - 1$

Divide each term in $- 3 y^{2} = x - 1$ by $- 3$ and simplify.

Divide each term in $- 3 y^{2} = x - 1$ by $- 3$ .

$\frac{- 3 y^{2}}{- 3} = \frac{x}{- 3} + \frac{- 1}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y^{2}}{- 3} = \frac{x}{- 3} + \frac{- 1}{- 3}$

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

$y^{2} = \frac{x}{- 3} + \frac{- 1}{- 3}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y^{2} = - \frac{x}{3} + \frac{- 1}{- 3}$

Dividing two negative values results in a positive value.

$y^{2} = - \frac{x}{3} + \frac{1}{3}$

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

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

Simplify $\pm \sqrt{- \frac{x}{3} + \frac{1}{3}}$ .

Combine the numerators over the common denominator.

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

Rewrite $\sqrt{\frac{- x + 1}{3}}$ as $\frac{\sqrt{- x + 1}}{\sqrt{3}}$ .

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

Multiply $\frac{\sqrt{- x + 1}}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt{- x + 1}}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

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

Raise $\sqrt{3}$ to the power of $1$ .

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

Raise $\sqrt{3}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Combine using the product rule for radicals.

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

Reorder factors in $\pm \frac{\sqrt{(- x + 1) \cdot 3}}{3}$ .

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

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.

$y = \frac{\sqrt{3 (- x + 1)}}{3}$

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

$y = - \frac{\sqrt{3 (- x + 1)}}{3}$

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

$y = \frac{\sqrt{3 (- x + 1)}}{3}$

$y = - \frac{\sqrt{3 (- x + 1)}}{3}$

$y = \frac{\sqrt{3 (- x + 1)}}{3}$

$y = - \frac{\sqrt{3 (- x + 1)}}{3}$

$y = \frac{\sqrt{3 (- x + 1)}}{3}$

$y = - \frac{\sqrt{3 (- x + 1)}}{3}$

Step 3

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

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

Step 4

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

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

Find the range of $f (x) = 1 - 3 x^{2}$ .

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

Interval Notation:

$(- \infty, 1]$

Find the domain of $\frac{\sqrt{3 (- x + 1)}}{3}$ .

Set the radicand in $\sqrt{3 (- x + 1)}$ greater than or equal to $0$ to find where the expression is defined.

$3 (- x + 1) \geq 0$

Solve for $x$ .

Divide each term in $3 (- x + 1) \geq 0$ by $3$ and simplify.

Divide each term in $3 (- x + 1) \geq 0$ by $3$ .

$\frac{3 (- x + 1)}{3} \geq \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (- x + 1)}{3} \geq \frac{0}{3}$

Divide $- x + 1$ by $1$ .

$- x + 1 \geq \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$- x + 1 \geq 0$

Subtract $1$ from both sides of the inequality.

$- x \geq - 1$

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x \leq 1$

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

$(- \infty, 1]$

Find the domain of $f (x) = 1 - 3 x^{2}$ .

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

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

Step 5

Find the Inverse y=1-3x^2

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