# Homework Help: Composite functions

1. Apr 15, 2006

### F.B

This is the only section so far that i dont understand, i don't know how to

1. Let g(x)= x-3. Find a function of f so that f(g(x)) = x^2

2. Let f(x) = x^2. Find a function g so that f(g(x)) = x^2 + 8x + 16

3. Let f(x) = x + 4 and g(x) = (x-2)^2. Find a function u so that
f(g(u(x))) = 4x^2 - 8x + 8

4. Let y = f(x^2 + 3x - 5). Find y' when x = 1, given that f'(-1) = 2

2. Apr 15, 2006

### GregA

in your first one you are given g(x) = x-3 ...you then have to find a function f of x such that f(x-3) in someway maps to x^2...as a little example:
suppose g(x) = x+1 and you need to find f(x) such that f(g(x))=x^3

Firstly lets find $$(x+1)^3$$...we get:
$$x^3 +3x^2 +3x +1$$... and this is $$3x^2 +3x +1$$ too much!...now lets try and get rid of it...if we say $$f(g(x))=x^3-3x^2$$ we end up with...
$$(x^3 +3x^2 +3x +1) - 3(x^2 + 2x +1)$$...and after simplifying we have $$x^3 - 3x -2$$...still $$-1(3x+2)$$ too much...
so now lets say $$f(g(x)) = x^3-3x^2 +3x$$...once you do the arithmetic you need to somehow get rid of 1...see whats happening? can you apply this methodology (or a better one) to your questions?

*edit in the 4th question...(didn't look at it carefully enough at my first visit) you can imagine that $$f(x^2 + 3x - 5) = f(g(h(x))$$ where $$g(x) = x^2 + 3x - 5$$ and h(x) is some other function of x...try to find h(x) such that it's effect on f(g(h(x))) is just enough that when you differentiate the result you have the correct terms in x to make $$f'(-1) = 2$$

(ie: if you decided to let $$h(x) = x^3$$ you'd have a serious case over over-kill because f(g(h(x))) would be $$x^6 +3x^3 - 5$$... and f'(x) here is gonna be miles away from 2 when x = -1 )

one more thing...finish up by then finding f'(1)

(edit2...apologies for what may be an un-wieldy reply...I might need to practice saying what I mean without going round the houses )

Last edited: Apr 15, 2006
3. Apr 15, 2006

### VietDao29

Let g(x)= x-3. Find a function of f so that f(g(x)) = x^2
Saying f(g(x)) = x2 is the same to saying that f(x - 3) = x2, right? Since g(x) = x - 3.
This problem means that you should find a function is terms of (x - 3), such that the result is x2.
Okay, I think I'll give you another example
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Example:
Let g(x) = x - 5
Find f such that f(g(x)) = x2 + x
f(g(x)) should be a quadratic function, and g(x) is a linear one. So to get a second degree polynomial from g(x), we musy first square it, right?
Let's square it:
(x - 5)2 = x2 - 10x + 25
Now you have x2, but you only need x2 + x, not x2 - 10x + 25, right?
So x2 - 10x + 25 = x2 + x - 11x + 25 (the -11x + 25, you don't need, right?)
So we must try to eliminate that, first, try to eliminate the -11x part, using (x - 5). We must add another 11(x - 5), right?
So:
(x - 5)2 + 11(x - 5) = x2 + x - 11x + 25 + 11x - 55
= x2 + x - 30.
Now to get x2 + x, we need to add to that expression a constant 30, right?
So we have:
(x - 5)2 + 11(x - 5) + 30 = x2 + x.
Exactly what we want to find.
So f(x - 5) = (x - 5)2 + 11(x - 5) + 30
Let y = x - 5
So f(y) = y2 + 11y + 30.
That the function f you need to find.
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For problem #2,
f(g(x)) = g(x)2, right?
f(g(x)) = x2 + 8x + 16 = (...)2
What should be (...), you think? Can you find g(x) from here?
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Let's see if you can do problem #3 on your own.
If you are stuck somewhere, just shout it out. :)
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You know the chain rule right?
$$\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$$
So:
$$\frac{dy}{dx} = \frac{dy}{d(x ^ 2 + 3x - 5)} \times \frac{d(x ^ 2 + 3x - 5)}{dx} = f'(x ^ 2 + 3x - 5) \times (2x + 3)$$
Can you get this?
What's the output of x2 + 3x - 5 at x = 1?
And you also know that f'(-1) = 2.
Can you go from here?
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Can you get this? Is there anything left unclear? :)

Last edited: Apr 15, 2006