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need math wiz to help find the inverse function

R

RayEarth

Senior member
i need to find the inverse function of: f(x)=x^3+x+1

i tried factoring an x to get: x(x^2+1)=y-1 and then i divided by x to get y-1 over x and moved everything to one side to get:

x^2 - 1/x * (y-1) + 1=0 then i tried using the quadratic formula using a=1 b=-(y-1) and tried c=1 but it didn't work, c=0 didn't help either.

i need steps for solving any function with x^3, i only remember how to work with x^2.
 
I guess i'm losing it...😛


I only got this far

y=x^3+x+1
y-1=x^3+x
y-1=x(x^2 + 1)


(y-1)/x = x^2 + 1
[(y-1)/x] -1 = x^2
(y-1) - (1/x) =x
y-1= x - (1/x)
 
Are you sure the question isn't find g'(a) where g is the inverse function of f(x)=x^3+x+1? To solve that you don't actually need to know what g actually is.
 
you're right Kyteland, there's is a thm for that, and i'm not sure if i need to switch x and y after we finish if i use this thm

here's the thm for this section: g'(a)=1/f'(g(a))

it asks to find f^-1'(a) and a=1

i don't know where and when to apply the the theorem.
 
So you have this theorem: g'(a)=1/f'(g(a))

you also know f'(x) = 3*x^2+1. This is easy to get because you are given f(x)

Since a=1 that gives you g'(1) = 1/(3*(g(1))^2+1)

so all you really have to solve for is g(1).

since g() is the inverse function of f() you know that for any g(a)=b that f(b)=a. Since you want to solve for g(1) solve the equation f(x)=1

f(x) = 1 = x^3+x+1
0 = x^3+x
0 = x*(x^2+1)
x = 0, +or- i

so x=0 which means that g(1) = 0. Now plug that back in to the original formula.

g'(1) = 1/(3*(g(1))^2+1)
= 1/(3*0^2+1)
= 1/1
= 1

edit: Also I hope you realize that when you write f^-1() that doesn't actually mean 1/f(). That notation means the inverse of the function, not to take it to the power of -1

so if f(x) = x^3 then f^-1(x) = x^(1/3). It does not mean that it is 1/x^3
 
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