Find the Inverse y=5x^2+10

$y = 5 x^{2} + 10$

Step 1

Interchange the variables.

$x = 5 y^{2} + 10$

Step 2

Solve for $y$ .

Rewrite the equation as $5 y^{2} + 10 = x$ .

$5 y^{2} + 10 = x$

Subtract $10$ from both sides of the equation.

$5 y^{2} = x - 10$

Divide each term in $5 y^{2} = x - 10$ by $5$ and simplify.

Divide each term in $5 y^{2} = x - 10$ by $5$ .

$\frac{5 y^{2}}{5} = \frac{x}{5} + \frac{- 10}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y^{2}}{5} = \frac{x}{5} + \frac{- 10}{5}$

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

$y^{2} = \frac{x}{5} + \frac{- 10}{5}$

Simplify the right side.

Divide $- 10$ by $5$ .

$y^{2} = \frac{x}{5} - 2$

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

$y = \pm \sqrt{\frac{x}{5} - 2}$

Simplify $\pm \sqrt{\frac{x}{5} - 2}$ .

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$y = \pm \sqrt{\frac{x}{5} - 2 \cdot \frac{5}{5}}$

Combine $- 2$ and $\frac{5}{5}$ .

$y = \pm \sqrt{\frac{x}{5} + \frac{- 2 \cdot 5}{5}}$

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{x - 2 \cdot 5}{5}}$

Multiply $- 2$ by $5$ .

$y = \pm \sqrt{\frac{x - 10}{5}}$

Rewrite $\sqrt{\frac{x - 10}{5}}$ as $\frac{\sqrt{x - 10}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{x - 10}}{\sqrt{5}}$

Multiply $\frac{\sqrt{x - 10}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{x - 10}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{x - 10}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{\sqrt{5} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{\sqrt{5}}^{1} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{\sqrt{5}}^{1} {\sqrt{5}}^{1}}$

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

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{\sqrt{5}}^{1 + 1}}$

Add $1$ and $1$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{\sqrt{5}}^{2}}$

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{(5^{\frac{1}{2}})}^{2}}$

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

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5^{\frac{1}{2} \cdot 2}}$

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

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5^{\frac{2}{2}}}$

Rewrite the expression.

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5^{1}}$

Evaluate the exponent.

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5}$

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{(x - 10) \cdot 5}}{5}$

Reorder factors in $\pm \frac{\sqrt{(x - 10) \cdot 5}}{5}$ .

$y = \pm \frac{\sqrt{5 (x - 10)}}{5}$

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.

$y = \frac{\sqrt{5 (x - 10)}}{5}$

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

$y = - \frac{\sqrt{5 (x - 10)}}{5}$

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

$y = \frac{\sqrt{5 (x - 10)}}{5}$

$y = - \frac{\sqrt{5 (x - 10)}}{5}$

$y = \frac{\sqrt{5 (x - 10)}}{5}$

$y = - \frac{\sqrt{5 (x - 10)}}{5}$

$y = \frac{\sqrt{5 (x - 10)}}{5}$

$y = - \frac{\sqrt{5 (x - 10)}}{5}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ is the inverse of $f (x) = 5 x^{2} + 10$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 5 x^{2} + 10$ and $f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ and compare them.

Find the range of $f (x) = 5 x^{2} + 10$ .

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

Interval Notation:

$[10, \infty)$

Find the domain of $\frac{\sqrt{5 (x - 10)}}{5}$ .

Set the radicand in $\sqrt{5 (x - 10)}$ greater than or equal to $0$ to find where the expression is defined.

$5 (x - 10) \geq 0$

Solve for $x$ .

Divide each term in $5 (x - 10) \geq 0$ by $5$ and simplify.

Divide each term in $5 (x - 10) \geq 0$ by $5$ .

$\frac{5 (x - 10)}{5} \geq \frac{0}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 (x - 10)}{5} \geq \frac{0}{5}$

Divide $x - 10$ by $1$ .

$x - 10 \geq \frac{0}{5}$

Simplify the right side.

Divide $0$ by $5$ .

$x - 10 \geq 0$

Add $10$ to both sides of the inequality.

$x \geq 10$

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

$[10, \infty)$

Find the domain of $f (x) = 5 x^{2} + 10$ .

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 $f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ is the range of $f (x) = 5 x^{2} + 10$ and the range of $f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ is the domain of $f (x) = 5 x^{2} + 10$ , then $f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ is the inverse of $f (x) = 5 x^{2} + 10$ .

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$

Step 5

Find the Inverse y=5x^2+10

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