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

Question

anonymous · Answer

$$y=f(x)$$
$$y = 5x+1$$
$$\frac{y-1}{5}= x$$
Hence 

$$f^{-1}(x) = \frac{x-1}{5}$$

angela210793 · Answer

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

anonymous · Answer

To find $f^{-1}(x)$ remember that $f(f^{-1}(x)) = x$

Therefore:

$$f(f^{-1}(x)) = 5(f^{-1}(x)) +1$$$$\implies x = 5f^{-1}(x) + 1$$$$\implies x-1 = 5(f^{-1}(x))$$$$\implies f^{-1}(x) = \frac{1}{5}(x-1)$$

anonymous · Answer

Or at least that's how I remember

anonymous · Answer

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.