Find the Inverse f(x)=x^5-3

$f (x) = x^{5} - 3$

Step 1

Write $f (x) = x^{5} - 3$ as an equation.

$y = x^{5} - 3$

Step 2

Interchange the variables.

$x = y^{5} - 3$

Step 3

Solve for $y$ .

Rewrite the equation as $y^{5} - 3 = x$ .

$y^{5} - 3 = x$

Add $3$ to both sides of the equation.

$y^{5} = x + 3$

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

$y = \sqrt[5]{x + 3}$

Step 4

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

$f^{- 1} (x) = \sqrt[5]{x + 3}$

Step 5

Verify if $f^{- 1} (x) = \sqrt[5]{x + 3}$ is the inverse of $f (x) = x^{5} - 3$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (x^{5} - 3)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (x^{5} - 3) = \sqrt[5]{(x^{5} - 3) + 3}$

Add $- 3$ and $3$ .

$f^{- 1} (x^{5} - 3) = \sqrt[5]{x^{5} + 0}$

Add $x^{5}$ and $0$ .

$f^{- 1} (x^{5} - 3) = \sqrt[5]{x^{5}}$

Pull terms out from under the radical, assuming real numbers.

$f^{- 1} (x^{5} - 3) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\sqrt[5]{x + 3})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[5]{x + 3}) = {(\sqrt[5]{x + 3})}^{5} - 3$

Rewrite ${\sqrt[5]{x + 3}}^{5}$ as $x + 3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{x + 3}$ as ${(x + 3)}^{\frac{1}{5}}$ .

$f (\sqrt[5]{x + 3}) = {({(x + 3)}^{\frac{1}{5}})}^{5} - 3$

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

$f (\sqrt[5]{x + 3}) = {(x + 3)}^{\frac{1}{5} \cdot 5} - 3$

Combine $\frac{1}{5}$ and $5$ .

$f (\sqrt[5]{x + 3}) = {(x + 3)}^{\frac{5}{5}} - 3$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\sqrt[5]{x + 3}) = {(x + 3)}^{\frac{5}{5}} - 3$

Rewrite the expression.

$f (\sqrt[5]{x + 3}) = (x + 3) - 3$

Simplify.

$f (\sqrt[5]{x + 3}) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f (\sqrt[5]{x + 3}) = x + 0$

Add $x$ and $0$ .

$f (\sqrt[5]{x + 3}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[5]{x + 3}$ is the inverse of $f (x) = x^{5} - 3$ .

$f^{- 1} (x) = \sqrt[5]{x + 3}$

Find the Inverse f(x)=x^5-3

Related Questions

Support LCMGCF.com by disabling your adblocker.