SOLUTION: How can I multiply a inverse function by an other inverse function

Question 1205142: How can I multiply a inverse function by an other inverse function

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the same way you would multiply any equation by any equation.
for example:
equation 1 is y = x^2
equation 2 is y = x^3
inverse equation 1 is y = x^(1/2)
inverse equation 2 is y = x^(1/3)
inverse equation 1 by inverse equation 2 to get y = x^(1/2) * x^(1/3) = x^(1/6)
it looks like if you multiply the original equations by each other you would get y = x^6 and the inverse equation to that is y = x^(1/6).

so, i would say, you can probably do it 2 ways.
1.
multiply the inverse equations by each other as you would multiply any equation by each other.
2.
multiply the original equations by each other and then get the inverse of that.
if option 2 is true, then you can check that you did the multiplication correctly by doing it both ways and compare your answer to see if they are the same.

try it out and let me know how you do.
send me the original equations and the inverse equation as well and i'll give it a shot from this end.

RELATED QUESTIONS
inverse... (answered by Fombitz)

Does every function have an inverse function? If so, is the inverse function also a... (answered by josgarithmetic)

Determine if the following function has an inverse function. k(x)=3x^2−5 A. k has (answered by MathLover1)

How can i find the inverse of the function f(x) = x^3 -5 / x... (answered by stanbon)

(a) Let f(x): (-inf, 0) U (0 inf) --> R be defined by f(x) = x- (1/x) Show that f(x) has (answered by ikleyn)

I need help with this question. The inverse of an exponential is a blank... (answered by Edwin McCravy)

Graph the inverse function (answered by jim_thompson5910)

(a) Let f : (-\infty,0) \cup (0,\infty) \to \mathbb{R} be defined by f(x) = x -... (answered by math_tutor2020,ikleyn)

How do you find the inverse of a function? Create a sample function and find it's... (answered by josgarithmetic)