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

Replace with y: To find the inverse function, we start by replacing f(x) with y: y = (root(5)(x) + 1) / 5Swap x and y: Next, we swap x and y to begin solving for the new y, which will be our f^(-1)(x): x = (root(5)(y) + 1) / 5Eliminate fraction: Multiply both sides by 5 to eliminate the fraction: 5x = root(5)(y) + 1Isolate fifth root term: Subtract 1 from both sides to isolate the fifth root term: 5x - 1 = root(5)(y)Eliminate fifth root: Raise both sides to the power of 5 to eliminate the fifth root: (5x - 1)^5 = yInverse function: Now we have the inverse function: f^(-1)(x) = (5x - 1)^5

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

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

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

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

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

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

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Math Problems

Algebra 1

Multiplication with rational exponents

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Q. For the function

f(x)=\frac{\sqrt[5]{x}+1}{5}

, find

f^{-1}(x)

\newline

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

\newline

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

\newline

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

\newline

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

Replace with $y$ : To find the inverse function, we start by replacing $f(x)$ with $y$ : $y = \frac{\sqrt[5]{x} + 1}{5}$
Swap x and y: Next, we swap x and y to begin solving for the new y, which will be our $f^{-1}(x)$ : $x = \frac{\sqrt[5]{y} + 1}{5}$
Eliminate fraction: Multiply both sides by $5$ to eliminate the fraction: $5x = \sqrt[5]{y} + 1$
Isolate fifth root term: Subtract $1$ from both sides to isolate the fifth root term: $\newline$ $5x - 1 = \sqrt[5]{y}$
Eliminate fifth root: Raise both sides to the power of $5$ to eliminate the fifth root: $(5x - 1)^5 = y$
Inverse function: Now we have the inverse function: $f^{-1}(x) = (5x - 1)^5$