For the function f(x)=7x^((1)/(5)), find f^(-1)(x). f^(-1)(x)=(7x)^(5) f^(-1)(x)=(x^(5))/(7) f^(-1)(x)=((x)/(7))^(5) f^(-1)(x)=(x^((1)/(5)))/(7)

Rewrite with y: 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. Let's start by rewriting the function with y instead of f(x): y = 7x^(1/5)Interchange x and y: Now, interchange x and y to find the inverse function: x = 7y^(1/5)Isolate y: To solve for y, we need to isolate y on one side of the equation. First, we'll raise both sides of the equation to the power of 5 to get rid of the exponent on the right side: x^5 = (7y^(1/5))^5Simplify right side: Simplifying the right side of the equation by raising both the base and the exponent to the power of 5, we get: x^5 = 7^5 * yDivide by constant: Now, divide both sides of the equation by 7^5 to solve for y: y = x^5 / 7^5Final inverse function: Since 7^5 is just a constant, we can simplify the expression by writing it as y = (x^5) / (7^5). However, 7^5 is a large number, and it's more convenient to leave it as 7 to indicate the operation that needs to be performed. So, the inverse function is: f^(-1)(x) = (x^5) / 7

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For the function
f(x)=7x^((1)/(5)), find
f^(-1)(x).

f^(-1)(x)=(7x)^(5)

f^(-1)(x)=(x^(5))/(7)

f^(-1)(x)=((x)/(7))^(5)

f^(-1)(x)=(x^((1)/(5)))/(7)

For the function $f(x)=7 x^{\frac{1}{5}}$ , find $f^{-1}(x)$ . $\newline$ $f^{-1}(x)=(7 x)^{5}$ $\newline$ $f^{-1}(x)=\frac{x^{5}}{7}$ $\newline$ $f^{-1}(x)=\left(\frac{x}{7}\right)^{5}$ $\newline$ $f^{-1}(x)=\frac{x^{\frac{1}{5}}}{7}$

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Math Problems

Algebra 1

Multiplication with rational exponents

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Q. For the function

f(x)=7 x^{\frac{1}{5}}

, find

f^{-1}(x)

\newline

f^{-1}(x)=(7 x)^{5}

\newline

f^{-1}(x)=\frac{x^{5}}{7}

\newline

f^{-1}(x)=\left(\frac{x}{7}\right)^{5}

\newline

f^{-1}(x)=\frac{x^{\frac{1}{5}}}{7}

Rewrite with $y$ : 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 $y$ . Let's start by rewriting the function with $y$ instead of $f(x)$ : $\newline$ $y = 7x^{\frac{1}{5}}$
Interchange x and y: Now, interchange x and y to find the inverse function: $x = 7y^{\frac{1}{5}}$
Isolate y: To solve for y, we need to isolate y on one side of the equation. First, we'll raise both sides of the equation to the power of $5$ to get rid of the exponent on the right side: $\newline$ $x^5 = (7y^{1/5})^5$
Simplify right side: Simplifying the right side of the equation by raising both the base and the exponent to the power of $5$ , we get: $\newline$ $x^5 = 7^5 \times y$
Divide by constant: Now, divide both sides of the equation by $7^5$ to solve for $y$ : $\newline$ $y = \frac{x^5}{7^5}$
Final inverse function: Since $7^5$ is just a constant, we can simplify the expression by writing it as $y = \frac{x^5}{7^5}$ . However, $7^5$ is a large number, and it's more convenient to leave it as $7$ to indicate the operation that needs to be performed. So, the inverse function is: $\newline$ $f^{-1}(x) = \frac{x^5}{7}$