finding inverse functions
I'm having trouble finding inverse functions. Does anyone have any general tips? Or more specfically problems like 13 and 17 in the homework from section 3.7.
posted by Carrie Desmond at 4:26 PM
Blog for students in Mr. Karafiol's Period 4 Precalculus BC class to talk about math, learning math, etc.
posted by Carrie Desmond at 4:26 PM
3 Comments:
i am also, it is especially difficult for me to find the inverse function when the variable is in both numerator and denominator of the expression. anyone know how to find them?
To find the inverse of a function replace f(x) with y and simply solve for x in terms of y. For functions with x in both the numerator and the denominator you would take the same approach.
For Carrie
3.7.17
f(x)=1/x--->y=1/x
xy=1--->x=1/y
f^-1(x)=1/x
For Taylor
3.7.20
f(x)=x/(x+1)--->y=x/(x+1)
y(x+1)=x--->xy+y=x
xy-x=-y--->x(y-1)=-y--->x=-y/(y-1)
f^-1(x)=-x/(x-1)
You can also use the solve key in your calculator if you don't want to do the algebra.
The key point is that an equation like xy + x = y can be solved for y by subtracting xy from both sides, then factoring: x = y - xy = y(1 - x). Then y = x/(1-x).
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