If $$f(x)=5x^3-25x+9$$ Find $$f^-1(f(3))$$ Note: That is supposed to be the inverse function of f. If $$f(x)=5x^3-25x+9$$ Find $$f^-1(f(3))$$ Note: That is supposed to be the inverse function of f. @Mathematics

Question

amistre64 · Answer

the general consensus for finding an inverse, or something akin to an inverse is to swap out x for y and resolve for y

anonymous · Answer

What would do in this case?

amistre64 · Answer

$$y=5x^3-25x+9$$
$$x=5y^3-25y+9$$

anonymous · Answer

Do you put in 3 before or after though?

amistre64 · Answer

but in this case, since the inverse of a function undoes the function; you get back to where you started

amistre64 · Answer

$$f^{-1}(f(x))=x$$

amistre64 · Answer

since x=3; the answer would just undo itself back to 3

anonymous · Answer

So you would not even have to plug in 3 for x?

amistre64 · Answer

nope, since you already know where you started at: x=3, and you know that an inverse simply gets you back to where you started from.

amistre64 · Answer

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amistre64 · Answer

f(3) = some number ...

f-1(some number) = 3

amistre64 · Answer

f-1(f(3))

f-1(some number)

amistre64 · Answer

$$x=5y^3-25y+9$$
$$x-9=5y^3-25y$$
$$x-9=5y(y^2-5)$$
$$\frac{x-9}{5}=y(y^2-5)$$
$$\frac{x-9}{5}=y(y-\sqrt{5})(y+\sqrt{5})$$
yeah, thats gonna get messy

best bet is to just use the definition of an inverse :)

jamesj · Answer

Actually, you have to be a little more careful.

Suppose the function were g(x) = x^2 and the question asked you what is the value of
$$ g^{-1}(g(1))$$

Well, $ g(1) = 1^2 = 1$.  But $g^{-1}(1)$ has two values, $ \{+1, -1\} $

Hence, for your problem, you can be certain that 3 is one value of $ f^{-1}(f(3))$.  But you should check for others.

jamesj · Answer

The way to do that is fairly straightforward
1. First calculate f(3)
2. Then find all x such f(x) = f(3)

The solutions to step 2 are all the possible values of $ f^{-1}(f(3))$.

In my example above step 2 is finding all x such that g(x) = 1.  
I.e., x such that $ x^2 = 1 $.  That's what gave us the two values, +1 and -1.

amistre64 · Answer

I would contend that in this case, they give you the intial value of x=3 so that you dont have to undo a spurious result.

if it had been stated such that:  f(x) = x^2,  find f-1(9) there would be some issue as to whether or not the intended results should be 3 or -3

but, given that you know the "x" from which it came from; f(x)=x^2, find f-1(f(3)) would simply result in 3.

is that correct?

jamesj · Answer

No, it's not because of the order of operations.  With your last version of the problem
f(3) = 9, but $ f^{-1}(9) $ still has two values.

The order is important because instead the question were find $f(f^{-1}(a))$, then provided a was in the range of f originally, $f(f^{-1}(a))$ is uniquely equal to a, and that's because f itself is a function.  But the relation $f^{-1}$ is not.

jamesj · Answer

this is one of the reasons we got to great pains to define "principle values" of Arcsin, Arccos etc in such a way as they are functions.  But without having done so, expressions such as
$$ \sin^{-1}(0) $$
has an infinite number of values.

jamesj · Answer

I might as well finish solving this problem and point something else out.

With the original definition of f, f(x) = 5x^3 - 25x + 9, f(3) = 69.

Now what are the solutions of
f(x) = 69?

     5x^3 - 25x + 9 = 69
iff    x^3 -  5x + 12 = 0

We know x = 3 is already a root of this equation so we can factor out x-3 to obtain

$$ (x-3)( x^2 + 3x + 4) = 0 $$

Now are there any real x such that $ x^2 + 3x + 4 = 0 $?  No.

Hence in fact the only (real) x such that f(x) = f(3) is x = 3.

So this might seem like wasted effort.  But it wasn't, because we didn't know before the fact that this was the case.  So you can carry this method in to other problems of this sort in the future.

amistre64 · Answer

that, was a very good answer :)