Find the Inverse e^(3x^2)

$e^{3 x^{2}}$

Step 1

Interchange the variables.

$x = e^{3 y^{2}}$

Step 2

Solve for $y$ .

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

$e^{3 y^{2}} = x$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{3 y^{2}}) = \ln (x)$

Expand the left side.

Expand $\ln (e^{3 y^{2}})$ by moving $3 y^{2}$ outside the logarithm.

$3 y^{2} \ln (e) = \ln (x)$

The natural logarithm of $e$ is $1$ .

$3 y^{2} \cdot 1 = \ln (x)$

Multiply $3$ by $1$ .

$3 y^{2} = \ln (x)$

Divide each term in $3 y^{2} = \ln (x)$ by $3$ and simplify.

Divide each term in $3 y^{2} = \ln (x)$ by $3$ .

$\frac{3 y^{2}}{3} = \frac{\ln (x)}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y^{2}}{3} = \frac{\ln (x)}{3}$

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

$y^{2} = \frac{\ln (x)}{3}$

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

$y = \pm \sqrt{\frac{\ln (x)}{3}}$

Simplify $\pm \sqrt{\frac{\ln (x)}{3}}$ .

Rewrite $\sqrt{\frac{\ln (x)}{3}}$ as $\frac{\sqrt{\ln (x)}}{\sqrt{3}}$ .

$y = \pm \frac{\sqrt{\ln (x)}}{\sqrt{3}}$

Multiply $\frac{\sqrt{\ln (x)}}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$y = \pm \frac{\sqrt{\ln (x)}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{\ln (x)}}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$y = \pm \frac{\sqrt{\ln (x)} \sqrt{3}}{\sqrt{3} \sqrt{3}}$

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

$y = \pm \frac{\sqrt{\ln (x)} \sqrt{3}}{3}$

Simplify the numerator.

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{\ln (x) \cdot 3}}{3}$

Reorder $\ln (x)$ and $3$ .

$y = \pm \frac{\sqrt{3 \cdot \ln (x)}}{3}$

Simplify $3 \cdot \ln (x)$ by moving $3$ inside the logarithm.

$y = \pm \frac{\sqrt{\ln (x^{3})}}{3}$

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.

$y = \frac{\sqrt{\ln (x^{3})}}{3}$

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

$y = - \frac{\sqrt{\ln (x^{3})}}{3}$

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

$y = \frac{\sqrt{\ln (x^{3})}}{3}$

$y = - \frac{\sqrt{\ln (x^{3})}}{3}$

$y = \frac{\sqrt{\ln (x^{3})}}{3}$

$y = - \frac{\sqrt{\ln (x^{3})}}{3}$

$y = \frac{\sqrt{\ln (x^{3})}}{3}$

$y = - \frac{\sqrt{\ln (x^{3})}}{3}$

Step 3

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

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

Step 4

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

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

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

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 $\frac{\sqrt{\ln (x^{3})}}{3}$ .

Set the argument in $\ln (x^{3})$ greater than $0$ to find where the expression is defined.

$x^{3} > 0$

Solve for $x$ .

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

$\sqrt[3]{x^{3}} > \sqrt[3]{0}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$x > \sqrt[3]{0}$

Simplify the right side.

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x > \sqrt[3]{0^{3}}$

Pull terms out from under the radical.

$x > 0$

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

$\ln (x^{3}) \geq 0$

Solve for $x$ .

Convert the inequality to an equality.

$\ln (x^{3}) = 0$

Solve the equation.

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x^{3})} = e^{0}$

Rewrite $\ln (x^{3}) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = x^{3}$

Solve for $x$ .

Rewrite the equation as $x^{3} = e^{0}$ .

$x^{3} = e^{0}$

Subtract $e^{0}$ from both sides of the equation.

$x^{3} - e^{0} = 0$

Simplify each term.

Anything raised to $0$ is $1$ .

$x^{3} - 1 \cdot 1 = 0$

Multiply $- 1$ by $1$ .

$x^{3} - 1 = 0$

Factor the left side of the equation.

Rewrite $1$ as $1^{3}$ .

$x^{3} - 1^{3} = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

$(x - 1) (x^{2} + x \cdot 1 + 1^{2}) = 0$

Simplify.

Multiply $x$ by $1$ .

$(x - 1) (x^{2} + x + 1^{2}) = 0$

One to any power is one.

$(x - 1) (x^{2} + x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x^{2} + x + 1 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x^{2} + x + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + x + 1$ equal to $0$ .

$x^{2} + x + 1 = 0$

Solve $x^{2} + x + 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

Multiply $2$ by $1$ .

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

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $+$ .

$x = \frac{- 1 + i \sqrt{3}}{2}$

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 + i \sqrt{3}}{2}$

Factor $- 1$ out of $i \sqrt{3}$ .

$x = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{2}$

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$x = \frac{- 1 (1 - i \sqrt{3})}{2}$

Move the negative in front of the fraction.

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

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $-$ .

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

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 - i \sqrt{3}}{2}$

Factor $- 1$ out of $- i \sqrt{3}$ .

$x = \frac{- 1 \cdot 1 - (i \sqrt{3})}{2}$

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$x = \frac{- 1 (1 + i \sqrt{3})}{2}$

Move the negative in front of the fraction.

$x = - \frac{1 + i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

The final solution is all the values that make $(x - 1) (x^{2} + x + 1) = 0$ true.

$x = 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Find the domain of $\ln (x^{3})$ .

Set the argument in $\ln (x^{3})$ greater than $0$ to find where the expression is defined.

$x^{3} > 0$

Solve for $x$ .

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

$\sqrt[3]{x^{3}} > \sqrt[3]{0}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$x > \sqrt[3]{0}$

Simplify the right side.

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x > \sqrt[3]{0^{3}}$

Pull terms out from under the radical.

$x > 0$

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

$(0, \infty)$

The solution consists of all of the true intervals.

$x \geq 1$

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

$[1, \infty)$

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

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.

$(- \infty, \infty)$

Find the range of the inverse.

Find the range of $\frac{\sqrt{\ln (x^{3})}}{3}$ .

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

Interval Notation:

$[0, \infty)$

Find the range of $- \frac{\sqrt{\ln (x^{3})}}{3}$ .

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

Interval Notation:

$(- \infty, 0]$

The union consists of all of the elements that are contained in each interval.

$(- \infty, \infty)$

Since the range of $f^{- 1} (x) = \frac{\sqrt{\ln (x^{3})}}{3}, - \frac{\sqrt{\ln (x^{3})}}{3}$ is not equal to the domain of $f (x) = e^{3 x^{2}}$ , then $f^{- 1} (x) = \frac{\sqrt{\ln (x^{3})}}{3}, - \frac{\sqrt{\ln (x^{3})}}{3}$ is not an inverse of $f (x) = e^{3 x^{2}}$ .

There is no inverse

Step 5

Find the Inverse e^(3x^2)

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