Question

If f(x) = 5x + 1 then its inverse function is given by f^-1(x) = ...?
Explainnn.

Answer 1

\[y=f(x)\]
\[y = 5x+1\]
\[\frac{y-1}{5}= x\]
Hence

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

Answer 2

x=5y+1
5y=x-1
y=(x-1)/5
f^-1=(x-1)/5

Answer 3

To find \(f^{-1}(x)\) remember that \(f(f^{-1}(x)) = x\)

Therefore:

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

Answer 4

Or at least that's how I remember

Answer 5

that makes absolutely no sence at all! I really don't understand f(x) at all. & when you add a ^-1 in there, it throws me off twice as much.