Find the Inverse f(x)=3x^2-1 , x>=0

$f (x) = 3 x^{2} - 1$ , $x \geq 0$

Step 1

Find the range of the given function.

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

$[- 1, \infty)$

Convert $[- 1, \infty)$ to an inequality.

$y \geq - 1$

Step 2

Find the inverse.

Interchange the variables.

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

Solve for $y$ .

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

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

Add $1$ to 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 $1$ .

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

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.

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

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

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

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

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

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

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 3

Find the inverse using the domain and the range of the original function.

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

Step 4

Find the Inverse f(x)=3x^2-1 , x>=0

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