SOLUTION: I'm working with a student myself at a Deaf school in Jamaica and we came across a problem that stumped me. The question is: If f(x)=2x-3, then find the value of f^-1(0) The

Algebra ->  Rational-functions -> SOLUTION: I'm working with a student myself at a Deaf school in Jamaica and we came across a problem that stumped me. The question is: If f(x)=2x-3, then find the value of f^-1(0) The       Log On


   

Question 166996: I'm working with a student myself at a Deaf school in Jamaica and we came across a problem that stumped me. The question is:
If f(x)=2x-3, then find the value of f^-1(0)
The part that stumps me is putting f to the power of -1.
We understand how to solve the problem if it's not but to a power.
Thank you so much for your help! I appreciate it greatly.
-Isaac
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Isaac, it's not a power but it stands for the inverse function.
For example, if I have a function f%28x%29 such that
f%281%29=5
then the inverse function, f%5E%28-1%29%28x%29 would be such that when I plug in x=5, I would get back y=1.
f%5E%28-1%29%285%29=1
.
.
.
1.To find the inverse, first change to x,y notation.
y=2%2Ax-3
Then interchange the positions of x and y and solve for y.
highlight%28x%29=sqrt%282%2Ahighlight%28y%29-3%29
x=2y-3
2y=x%2B3
y=%28x%2B3%29%2F2}
The solution for the new y is the inverse function.
f%5E%28-1%29%28x%29=%28x%2B3%29%2F2
.
.
.
Let's try it.
f%28x%29=2x-3
f%281%29=2%281%29-3=2-3=-1
Now when I plug in x=-1 into the inverse function,
I should get back y=1.
f%5E%28-1%29%28x%29=%28x%2B3%29%2F2
f%5E%28-1%29%28-1%29=%28-1%2B3%29%2F2
f%5E%28-1%29%28-1%29=2%2F2
f%5E%28-1%29%28-1%29=1
.
.
.
You interested at the value of the inverse function for x=0.
f%5E%28-1%29%28x%29=%28x%2B3%29%2F2
f%5E%28-1%29%280%29=%280%2B3%29%2F2
f%5E%28-1%29%280%29=3%2F2
You could have also solved it by asking what x makes my original function equal to zero.
2x-3=0
2x=3
x=3%2F2
The inverse functions interchange x and y values.
Hopefully that makes sense. Good luck!