Finding Inverse for (x+2)^2(x-3)

  • Thread starter Thread starter wegnes
  • Start date Start date
  • Tags Tags
    Inverse
AI Thread Summary
The function (x+2)^2(x-3) does not have an inverse for all real numbers because it is not one-to-one when graphed. However, it is possible to restrict the domain to intervals where the function is one-to-one, allowing for an inverse to exist. The discussion clarifies that the original question does not require finding the inverse but rather understanding the conditions under which it can exist. Identifying specific intervals where the function is one-to-one is crucial for determining valid restrictions. Overall, the key takeaway is that while the function lacks a global inverse, localized inverses can be established through domain restriction.
wegnes
Messages
6
Reaction score
0

Homework Statement



(x+2)^2(x-3) I am asked does an inverse exist for all real numbers? then I am asked can I restrict the domain and force and inverse to exist?

Homework Equations


I said no because the function is not one-to-one when graphed.


The Attempt at a Solution

I thought it I took the square root of the x's in the original equation then this would work.
 
Physics news on Phys.org
Okay, I'll warn you that if you don't clean this up you'll get reamed!
You're given a function. Let's call it f(x) and define it to be equal to (x+2)^2(x-3)
Now you're asked to take the inverse. Do that. There is one REAL value of x for which the inverse is undefined. This should be poignantly obvious. Part two of that question will make sense once you figure out part 1.
 
I'm afraid BrendanH has misunderstood the problem. You are not asked to take the inverse and are a whole lot more than ONE value at which that does not have an inverse!

wegnes, you say you have graphed it and know that it does not have an inverse because it is not one-to-one. Excellent! Now, on what intervals is the function one-to-one? There are several of those and so several different ways to restrict the function so that the restiction does have an inverse.

Since the question does not ask you to actually find an inverse, it is not necessary to "take the square root" or anything like that.
 
Thank you! That what I wasn't doing! You are a big help, I apreciate it;o)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top