Question

Find f^-1 for the function f(x) = (1-41x)/x

Answer 1

You can find the path to the solution using this "trick"
\[y=\frac{1-41x}{x}\Rightarrow x'=\frac{1-41y'}{y'}\]
And solving for y'.

Answer 2

\(\bf f(x) = \cfrac{1-41x}{x} \implies \color{red}{y} = \cfrac{1-41\color{blue}{x}}{\color{blue}{x}}\\\quad \\
f^{-1}(x) =\color{blue}{ x} = \cfrac{1-41\color{red}{y}}{\color{red}{y}} \implies x = \cfrac{1}{y}- \cfrac{41y}{y} \)

you find the inverse function by "switching about" the variables

Answer 3

would i multiply both sides by y to get rid of the denominator?

Answer 4

keep in mind that \(\bf x = \cfrac{1}{y}- \cfrac{41\cancel{y}}{\cancel{y}} \implies x = \cfrac{1}{y} -41\)

Answer 5

would it be y^-1 = 1/(x+ 41)

Answer 6

yeap \(\bf x = \cfrac{1}{y}- \cfrac{41y}{y} \implies x = \cfrac{1}{y} -41 \implies y = \cfrac{1}{x+41}\)

Answer 7

Awesome, thanks! I have one more like this if you would please check my answer. Find f^-1 for the function f(x) = (x-9)^3 + 3.

Answer 8

I got \[\sqrt[3]{(x-3)} + 9\]