How to Find the Inverse of a Polynomial: Step-by-Step Guide

AI Thread Summary
To find the inverse of the polynomial function f(x) = 2x^3 + 5x + 3 at f^-1(1), the task is to determine the value of x that satisfies f(x) = 1. The discussion clarifies that the problem does not require finding the full inverse function, which would be complex, but rather just the specific value of x for which f(x) equals 1. Participants emphasize that understanding the relationship between f(x) and f^-1(y) is crucial, indicating that if f(x) = y, then x = f^-1(y). Ultimately, the focus is on evaluating f(1) to find the corresponding x value.
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Homework Statement



Let f(x) = 2x^3 + 5x + 3
Find the inverse at f^-1(x) = 1

Homework Equations


N/A

The Attempt at a Solution



The only way that I know how to solve inverses is by solving for X, then replacing it by Y. Then I supposed I would sub 1 into the inverted polynomial. However I'm not sure how to solve for X.
My attempt:
y - 3 = 2x^3 + 5x
(y - 3)/x = x(2x^2 + 5)

Then, hopelessness. Any info will be extremely helpful for my test this Friday. Thank you.
 
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Your statement is a bit unclear -- I'm interpreting you to say that you want to find x = f-1(1) , that is, what value of x gives f(x) = 1 ?

I don't think they want you to solve the cubic equation 2x^{3} + 5x + 3 = 1 (unless they taught you how to do so in your course). This doesn't come out nicely at all...

I'm wondering if they're asking that if you had the function y = f-1(x)*, what value of x would give y = 1 ? This is equivalent to asking what f(1) equals: the value of f(1) is the number that f-1(x) would "take back" to 1 .*whatever that is exactly -- it wouldn't be pretty; instead, we will work with what is called the "implicit function"
 
buddy think again;
the answer is f(1)...
 
Yes, as both dynamicsolo and abhishek ghos are saying, you are completely misunderstanding the problem. The problem does NOT as you to find the inverse function, which would be extremely complex. It only asks you to find the single value f^{-1}(1).

Use the fact that if f(x)= y then x= f^{-1}(y).
 
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
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