For the function f(x)=7x^((1)/(3)), find f^(-1)(x). f^(-1)(x)=((x)/(7))^(3) f^(-1)(x)=(x^(3))/(7) f^(-1)(x)=(7x)^(3) f^(-1)(x)=7x^(3)

Original Function: To find the inverse function, f^(-1)(x), we need to switch the roles of x and y in the original function and then solve for y. The original function is f(x) = 7x^(1/3), which we can write as y = 7x^(1/3).Switch Roles: Replace y with x and x with y to get the equation for the inverse function: x = 7y^(1/3).Isolate y: To solve for y, we need to isolate y on one side of the equation. Start by dividing both sides of the equation by 7 to get x/7 = y^(1/3).Eliminate Cube Root: Now, raise both sides of the equation to the power of 3 to eliminate the cube root on the right side: (x/7)^3 = (y^(1/3))^3.Final Inverse Function: Simplifying the right side, we get (x/7)^3 = y. This is the inverse function of the original function.

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For the function
f(x)=7x^((1)/(3)), find
f^(-1)(x).

f^(-1)(x)=((x)/(7))^(3)

f^(-1)(x)=(x^(3))/(7)

f^(-1)(x)=(7x)^(3)

f^(-1)(x)=7x^(3)

For the function $f(x)=7 x^{\frac{1}{3}}$ , find $f^{-1}(x)$ . $\newline$ $f^{-1}(x)=\left(\frac{x}{7}\right)^{3}$ $\newline$ $f^{-1}(x)=\frac{x^{3}}{7}$ $\newline$ $f^{-1}(x)=(7 x)^{3}$ $\newline$ $f^{-1}(x)=7 x^{3}$

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. For the function

f(x)=7 x^{\frac{1}{3}}

, find

f^{-1}(x)

\newline

f^{-1}(x)=\left(\frac{x}{7}\right)^{3}

\newline

f^{-1}(x)=\frac{x^{3}}{7}

\newline

f^{-1}(x)=(7 x)^{3}

\newline

f^{-1}(x)=7 x^{3}

Original Function: To find the inverse function, $f^{-1}(x)$ , we need to switch the roles of $x$ and $y$ in the original function and then solve for $y$ . The original function is $f(x) = 7x^{\frac{1}{3}}$ , which we can write as $y = 7x^{\frac{1}{3}}$ .
Switch Roles: Replace $y$ with $x$ and $x$ with $y$ to get the equation for the inverse function: $x = 7y^{\frac{1}{3}}$ .
Isolate y: To solve for y, we need to isolate y on one side of the equation. Start by dividing both sides of the equation by $7$ to get $\frac{x}{7} = y^{\frac{1}{3}}$ .
Eliminate Cube Root: Now, raise both sides of the equation to the power of $3$ to eliminate the cube root on the right side: $(\frac{x}{$7$})^$3$ = (y^{\frac{$1$}{$3$}})^$3$.
Final Inverse Function: Simplifying the right side, we get $(\frac{x}{7})^3 = y$. This is the inverse function of the original function.