find P^-1(P(x))

actresschanel said:
… how am I supposed to find P^-1(P(x))
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Have you seen the machine models, for a function and its inverse?

inverseFunction (Machine).JPG

The original input is the number x, and the output is the number f(x)

The inverse function runs that in reverse.

Put the number f(x) into the inverse function, and the output is the original number x.

[imath]\;[/imath]
 
Otis said:
Ah, I didn't realize that you were talking about a defined function. My first reply was a general answer.

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And a perfectly good one. But I am not sure that giving that definitional statement will clarify things for this students as well as working it out in detail.
 
actresschanel said:
I can find P^-1 (x) but how am I supposed to find P^-1(P(x))?

Thanks.
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If you have [math]P^{-1}(x)[/math] then sub in P(x) for x to get [math]P^{-1}(P(x))[/math].

So if [math]P(x) = -12x^2 + 40x + 200[/math] then everywhere in [math]P^{-1}(x)[/math] you have an x you need to put [math]-12x^2 + 40x + 200[/math].

-Dan
 
actresschanel said:
P(x) = -12x^2+40x+200
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If you are asking about the inverse of [MATH]P(x)=-12x^2+40x+200[/MATH], it does not exits.
To see that look at to plot of the function HERE.
There is no value such [MATH]P(x)=300[/MATH]So what definition of inverse function are you given?
 
Can we have the restriction on p(x)?
If you had read our posting guidelines it asks for the complete problem. Supplying the forum helpers with an incomplete problems just delays everyone from helping you.
 
Yes, we can find inverses on restricted domains of P(x). We could restrict the natural domain of P(x) to obtain new functions that always increase or always decrease (as x increases). Such functions are called 'one-to-one', and it's a necessary condition for an inverse function to exist.

Googling a phrase like how to find inverse of quadratic function will yield many online resources about this (video lectures and written lessons and examples).

Here is a basic video example.

So, the first step is to restrict the domain. If we're not given a restricted domain for P(x), then it might make sense to split the natural domain at the parabola's axis of symmetry: x=-b/(2a). Then we'll have a new function for the left half of the parabola [x from negative infinity to -b/(2a)] and a new function for the right half [x from -b/(2a) to positive infinity]. Each of those new functions has an inverse function, which may be found using the method shown at the link above (swapping symbols x and y, then solving for y).

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actresschanel said:
I can find P^-1 (x) but how am I supposed to find P^-1(P(x))?

Thanks.
Click to expand...

If a function [imath]f:A \rightarrow B[/imath] is invertible, means that [imath]f^{-1}: B \rightarrow A[/imath] exist and it's a function. Consider [imath]a\epsilon A[/imath] and [imath]b\epsilon B[/imath], then [imath]f(a)=b \Leftrightarrow f^{-1}(b)=a[/imath]. Now let's make the composition of the two functions: [math]f(f^{-1}(b))=f(a)=b[/math] Then, if [imath]P^{-1}[/imath] is the inverse function of [imath]P[/imath] [math]P(P^{-1}(x))=x[/math]
 
Eugenio said:
If a function [imath]f:A \rightarrow B[/imath] is invertible, means that [imath]f^{-1}: B \rightarrow A[/imath] exist and it's a function. Consider [imath]a\epsilon A[/imath] and [imath]b\epsilon B[/imath], then [imath]f(a)=b \Leftrightarrow f^{-1}(b)=a[/imath]. Now let's make the composition of the two functions: [math]f(f^{-1}(b))=f(a)=b[/math] Then, if [imath]P^{-1}[/imath] is the inverse function of [imath]P[/imath] [math]P(P^{-1}(x))=x[/math]
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But the above is true only if [math]f:A\to B[/math] is a bijection( one-to-one, onto). Otherwise you must adjust the domain and other adjustments as needed.
 
pka said:
But the above is true only if [math]f:A\to B[/math] is a bijection( one-to-one, onto). Otherwise you must adjust the domain and other adjustments as needed.
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The above is correct because the invertibility of f(x) is a hypothesis. A hypothesis doesn't need to be true.
 
f={ [imath]\big([/imath]x , f(x)[imath]\big)[/imath], x [imath]\in[/imath] D }, D is the Domain of f
f[imath]^{-1}[/imath]:={ [imath]\big([/imath]f(x) , x[imath]\big)[/imath], x [imath]\in[/imath] D } (f[imath]^{-1}[/imath] not necessarily a function)
So if f[imath]^{-1}[/imath] is a function, then f[imath]^{-1}[/imath](f(x))=x
 
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