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Calculus/Inverse Functions

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Expert: - 1/9/2009


Question
Hi, I'm having trouble solving 2 kinds of inverse functions, can you help me to approach this kind of problems and advice a little on how to solve them? Thanks.

The first kind is the equations with 2 polynomials. For example, (x-3)/(x+8). I have stopped learning math for sometime and am a little rusty on the properties and rules, please explain in a little more detail.

And the other one is polynomials with 1 or more exponents. For example, f(x)=100/(1+2^(-x)) and some functions with more than one exponent and polynomials together.

Answer
Take y = (x+3)/(x+8).

Multiply by x+8, giving xy + 8y = x + 3

Add -8y-x to both sides, giving xy-x = 3-8y.

Divide both sides by y-1, giving x = (3-8y)/(y-1).

That's my final answer.



For the next problem, note that ln2(x) is the (log of x) base 2.
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For this one, we take y = 100/(1+2^(-x)).

Multiply both sides by (1+2^(-x))/y, giving 1+2^(-x) = 100/y.

Subtract 1 from both sides, giving
  2^(-x) = (100/y) - 1 { note that 1 is y/y }.

Compute ln2 of both sides, giving -x = ln2((100-y)/y).

Multiply through by a negative, giving x = -ln2((100-y)/y).

Invert the fraction since a negative outside can be taken as an exponet, giving x = ln2(y/(100-y)).

A ln2(a/b) = ln2(a) - ln2(b), so our final answer is
x = ln2(y) - ln2(100-y).  

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