Find the Inverse 5x^2+1

$5 x^{2} + 1$

Step 1

Interchange the variables.

$x = 5 y^{2} + 1$

Step 2

Solve for $y$ .

Rewrite the equation as $5 y^{2} + 1 = x$ .

$5 y^{2} + 1 = x$

Subtract $1$ from both sides of the equation.

$5 y^{2} = x - 1$

Divide each term in $5 y^{2} = x - 1$ by $5$ and simplify.

Divide each term in $5 y^{2} = x - 1$ by $5$ .

$\frac{5 y^{2}}{5} = \frac{x}{5} + \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y^{2}}{5} = \frac{x}{5} + \frac{- 1}{5}$

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

$y^{2} = \frac{x}{5} + \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$y^{2} = \frac{x}{5} - \frac{1}{5}$

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

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

Simplify $\pm \sqrt{\frac{x}{5} - \frac{1}{5}}$ .

Combine the numerators over the common denominator.

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

Rewrite $\sqrt{\frac{x - 1}{5}}$ as $\frac{\sqrt{x - 1}}{\sqrt{5}}$ .

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

Multiply $\frac{\sqrt{x - 1}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt{x - 1}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{(x - 1) \cdot 5}}{5}$

Reorder factors in $\pm \frac{\sqrt{(x - 1) \cdot 5}}{5}$ .

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

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.

$y = \frac{\sqrt{5 (x - 1)}}{5}$

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

$y = - \frac{\sqrt{5 (x - 1)}}{5}$

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

$y = \frac{\sqrt{5 (x - 1)}}{5}$

$y = - \frac{\sqrt{5 (x - 1)}}{5}$

$y = \frac{\sqrt{5 (x - 1)}}{5}$

$y = - \frac{\sqrt{5 (x - 1)}}{5}$

$y = \frac{\sqrt{5 (x - 1)}}{5}$

$y = - \frac{\sqrt{5 (x - 1)}}{5}$

Step 3

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

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

Step 4

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

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

Find the range of $f (x) = 5 x^{2} + 1$ .

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 $\frac{\sqrt{5 (x - 1)}}{5}$ .

Set the radicand in $\sqrt{5 (x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$5 (x - 1) \geq 0$

Solve for $x$ .

Divide each term in $5 (x - 1) \geq 0$ by $5$ and simplify.

Divide each term in $5 (x - 1) \geq 0$ by $5$ .

$\frac{5 (x - 1)}{5} \geq \frac{0}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 (x - 1)}{5} \geq \frac{0}{5}$

Divide $x - 1$ by $1$ .

$x - 1 \geq \frac{0}{5}$

Simplify the right side.

Divide $0$ by $5$ .

$x - 1 \geq 0$

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 $f (x) = 5 x^{2} + 1$ .

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

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

Step 5

Find the Inverse 5x^2+1

Related Questions

Support LCMGCF.com by disabling your adblocker.