The given function f is differentiable. Verify that f has an inverse and find (f^(-1))'(c): f(x) = (x+3)/(x-1), x1; c=3.

Question

The given function f is differentiable. Verify that f has an inverse and find (f^(-1))'(c): f(x) = (x+3)/(x-1), x>1; c=3.

anonymous · Answer

Really. I'm just a little confused on how to verify that f has an inverse. Doesn't the function x have to always be increasing or decreasing in order to be one-to-one? With this function, it dips up and down (but never negative within the given parameters).

anonymous · Answer

i think this is one to one, lets check by finding the inverse

anonymous · Answer

oh to answer your question, the function does not have to be increasing or decreasing to be one to one unless it is continuous
this one is not

anonymous · Answer

Here is my work. I just don't know how I go about verifying the inverse.

anonymous · Answer

I did the second part of the problem (find (f^(-1))'(c) only.

anonymous · Answer

$$x=\frac{y+3}{y-1}$$
$$x(y-1)=y+3$$
$$xy-x=y+3$$
$$xy-y=x+3$$
$$y(x-1)=x+3$$
$$y=\frac{x+3}{x-1}$$ so this function not only has an inverse, it is its own inverse

anonymous · Answer

What are the rules in order for a function to be one-to-one? Does it always have to be increasing or decreasing, or does it just not have to be negative?

anonymous · Answer

you can use the formula you used, but you can also take the derivative and replace $x$ by $3$

anonymous · Answer

for a function $f$ to be one to one it means if $f(a)=f(b)$ then $a=b$

anonymous · Answer

you can prove the function is one to one by 
a) checking to see whether it passes the horizontal line test if you have the graph
b) putting $f(a)=f(b)$ and showing by algebra that $a=b$
c) finding an inverse that is a function 

since you can find the inverse, it must be one to one

anonymous · Answer

Ah, ok then. And what you did was step c then, right?

anonymous · Answer

yes, since now that you have the inverse, you can clearly find its derivative

anonymous · Answer

What kind of problem would you have run into if you COULDN'T find the inverse?

anonymous · Answer

Would you just not be able to solve for y? And that's how you'd know that there wasn't an inverse?

anonymous · Answer

well you might get something like $y=x\pm\sqrt{4-x^2}$ for example

anonymous · Answer

Ohh ok, since there would be more than one point on the graph, it wouldn't be one-to-one.

anonymous · Answer

or for a simpler example here is a simple not one to one function, $f(x)=x^2$
if you try to solve $y=x^2$ for $x$ you get $x=\pm\sqrt{y}$ so not a function

anonymous · Answer

Gotcha, thanks.