# Math Help - find the inverse function

1. ## find the inverse function

The question asks me to find the inverse of a function and find inverse of (3) and f(inverse of 2)

f(t)=t^5 + t^3 + t
I tried going by the steps, but it seemed weird.
y=t^5 + t^3 + t
t=y^5 + y^3 + y
So inverse of f(t) = t^5 + t^3 + t since I can't simplify if any further?

Then for inverse of (3), it would be 3^5 + 3^3 + 3 = 273?
For f(inverse of 2), it would be f(2^5 + 2^3 + 2) = f(42) = 42^5 + 42^3 + 42? Woah...but that seems kind of high...

2. ## Re: find the inverse function

Originally Posted by Taurus3
The question asks me to find the inverse of a function and find inverse of (3) and f(inverse of 2)

f(t)=t^5 + t^3 + t
I tried going by the steps, but it seemed weird.
y=t^5 + t^3 + t
t=y^5 + y^3 + y
So inverse of f(t) = t^5 + t^3 + t since I can't simplify if any further?

Then for inverse of (3), it would be 3^5 + 3^3 + 3 = 273?
For f(inverse of 2), it would be f(2^5 + 2^3 + 2) = f(42) = 42^5 + 42^3 + 42? Woah...but that seems kind of high...

The problem to find the inverse of a function of the type...

$y=f(t)= c_{1}\ t + c_{2}\ t^{2}+ c_{3}\ t^{3} + ...\ ;\ c_{1}\ne 0$

... today has been analysed here...

http://www.mathhelpforum.com/math-he...on-188961.html

Kind regards

$\chi$ $\sigma$

3. ## Re: find the inverse function

? Ouch. I still don't get it. Must you solve this as a complex function?

4. ## Re: find the inverse function

Originally Posted by Taurus3
? Ouch. I still don't get it. Must you solve this as a complex function?
Real and complex functions in this case are treated in the same fashion. The only You have to know are the basic derivation rules...

The alternative is to solve a quintic equation, for which a solving formula doesn't exist...

Kind regards

$\chi$ $\sigma$

5. ## Re: find the inverse function

Originally Posted by Taurus3
The question asks me to find the inverse of a function and find inverse of (3) and f(inverse of 2)

f(t)=t^5 + t^3 + t
I tried going by the steps, but it seemed weird.
y=t^5 + t^3 + t
t=y^5 + y^3 + y
So inverse of f(t) = t^5 + t^3 + t since I can't simplify if any further?

Then for inverse of (3), it would be 3^5 + 3^3 + 3 = 273?
For f(inverse of 2), it would be f(2^5 + 2^3 + 2) = f(42) = 42^5 + 42^3 + 42? Woah...but that seems kind of high...

I don't think you need the inverse function per se to answer the two questions,

for $f(t) = t^5 + t^3 + t$ , $f(1) = 3$ , so $f^{-1}(3) = 1$

for $f(t) = 2$

$0 = t^5 + t^2 + t - 2$

you may need to solve this w/ a calculator ... say you get $x = a$ , then $f^{-1}(2) = a$

edit: Since I've discovered that this post was originally in the calculus forum, I have to ask ... was this a calculus problem looking for the value of a derivative for an inverse function?