Solve Inverse Function of g(x)=(x2/e)+2 ln(x)-e

raptik
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Homework Statement


The function g(x) = (x2/e) + 2 ln(x) - e on (0,infinity) is one-to-one. Evaluate g-1(2)

Homework Equations


Find x in terms of y. Then switch x and y. Plug in 2 to the new equation.

The Attempt at a Solution


I can think of no way to get x explicitly in terms of y. I considered plugging in 2 to the original equation to get (x1,y1) and switching the two, but without a calculator, it seems unlikely to find a straightforward answer. Please help.
 
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The question only ask you to evaluate g^{-1}(2) it did not ask for a formula.

Think about what an inverse is if f(x) =y then f^{-1} (y) = ?.

Inverse "brings things back". If I throw a ball at you and "inverse" is you throwing the same ball at me, right ?
 
The answer choices are: A) 1 B) 2 C) e D) e2 E) 0

One of these is the correct "evaluation".

I follow that if f(x) =y then f-1 (y) = x

But when I don't know how to get to x, I'm not going to be able to solve this.
 
Have you considered finding g(e) ?
 
╔(σ_σ)╝ said:
Have you considered finding g(e) ?

Oh!

If I put x=e, then I get g(e) = 2. So (e,2). Then it's inverse is (2,e) which matches with g-1(2) to give me e. I see how that could work, but how did you have the intuition to add e to the original problem? I suppose it's a matter of plugging something that seems like it would give me the required value until it works. Thnx for help.
 
raptik said:
Oh!

If I put x=e, then I get g(e) = 2. So (e,2). Then it's inverse is (2,e) which matches with g-1(2) to give me e. I see how that could work, but how did you have the intuition to add e to the original problem? I suppose it's a matter of plugging something that seems like it would give me the required value until it works. Thnx for help.
When i looked at g i knew that the x^2 term would cause problems. I looked again at g and realized there was a 2 infront of lnx and i knew ln(e) =1. From there i realized that if i found g(e) i would get 2. ;-)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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