evil calculus problem - inverses

[bcnich]

Let f(x)=x³ + 3x - 1, calculate g'(3) when g'(x)= 1 /(f'(g(x))) and g is the inverse function for f.

i am so confused!

 

[Ishuda]

Let f(x)=x³ + 3x - 1, calculate g'(3) when g'(x)= 1 /(f'(g(x))) and g is the inverse function for f.

i am so confused!

Me too. What do you mean g is the inverse function of f? But, let's just play around. First
f'(x) = 3 x² + 3
which means
f'(g) = 3 g² + 3
and
\(\displaystyle g'=dg/dx=\frac{1}{3g^2+3}\)
Thus
\(\displaystyle (3g^2+3)dg=dx\)
and integrating each side we get
\(\displaystyle x=g^3+3g+c\)
Maybe that c is -1 and that's what you meant by g is the inverse function of f although that is not what I would expect it to mean.

I would normally expect g is the inverse function of f to mean the g(f(x)) = x. For example suppose f(x) = x+1 and g(x) = x-1, then
g(f(x)) = g(x+1) = (x+1) - 1 = x
Thus g is the inverse of f. Inverse functions are fairly easy to find for linear functions but for something like f(x)=x³ + 3x - 1, well I don't know.

 

[HallsofIvy]

So you want to evaluate \(\displaystyle \frac{1}{f'(g(3))}\).

The first thing you need to do is determine g(3), which is the real root to \(\displaystyle f(x)= x^3+ 3x- 1= 3\)
That turn out to be fairly simple- don't use a "formula" just evaluate f for some values of x.

 

[Ishuda]

So you want to evaluate \(\displaystyle \frac{1}{f'(g(3))}\).

The first thing you need to do is determine g(3), which is the real root to \(\displaystyle f(x)= x^3+ 3x- 1= 3\)
That turn out to be fairly simple- don't use a "formula" just evaluate f for some values of x.

Of course dum dum, if g is the inverse of f then f is the inverse of g so that f(g(3)) = 3 or \(\displaystyle f(g)= g^3+ 3g- 1= 3 \)

I see why you named yourself I shud a, yes you should have.

 

[HallsofIvy]

Of course dum dum, if g is the inverse of f then f is the inverse of g so that f(g(3)) = 3 or \(\displaystyle f(g)= g^3+ 3g- 1= 3 \)

I see why you named yourself I shud a, yes you should have.

I'm not sure who you are calling a "dum dum" or who you are referring to when you say "you named yourself". But what did you get for the value of g satisfying the equation above?

 

[Ishuda]

I'm not sure who you are calling a "dum dum" or who you are referring to when you say "you named yourself". But what did you get for the value of g satisfying the equation above?

I thought it was obvious - the name of the poster up above who said "Inverse functions are fairly easy to find for linear functions but for something like f(x)=x³ + 3x - 1, well I don't know." is named I shud a and yes, he should have seen that it was actually not a matter of finding the inverse to f but just finding the inverse at a single point. Oh, and g(3)=1 and a couple of others I didn't look for since (I think) we are dealing only with real numbers. Sorry if there was confusion about who shud a dun what

The reason for quoting HallsofIvy is that he (she?) is the one that made that clear [not that Ishuda should have seen that but that we only needed the g(x) which solved f(g(3)) = 3].