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CPhill

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Nombre de usuarioCPhill
Puntuación104899
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Preguntas 52
Respuestas 32587

 #5
avatar+104899 
+4
18 mar. 2014
 #1
avatar+104899 
+4
This kind of stuff - called "composite functions" - used to give me trouble, as well. I couldn't "see it" for awhile.

Let me see if I can help you understand it the way I finally saw it.

Suppose we had f(x) = 2x, and I wanted to know what f(2) was?? Well, x is like a "box" that's holding something inside. What is this something??

We could write f(x) = 2[ ] where the [ ] is the "box"......So. what are we putting in this box?? The answer is the "2" in f(2) !!!

So we have f(2) = 2[ ] = 2[2] = 4....and that's what f(2) "evaluates to!!

Okay....suppose we had f(x) = x^2 + 3x + 4, and I wanted to know what f(3) was??

So we have f(3) = [3]^2 + 3[3] + 4 = 9 + 9 + 4 = 22......so f(3) would evaluate to 22 .......note that, the "4" doesn't have any "box" associated with it because there is no "x" on this term !!!!

With me so far??

Now........ suppose instead that I was given two functions "f" and "g" so that f(x) = 3x and g(x) = 2x

We can rewrite these using our "boxes" as f(x) = 3[ ] and g(x) = 2[ ]

OK......here comes the "punchline"...so to speak!!

What would f(g) be?? Well...if f(3) meant that I was putting "3" into the box in "f,' then f(g) must mean that I'm putting the function "g" into the "box." in function "f" !!!!!

So we have f(g) =3[g] = 3[2x] = 6x........!!!

Alright....what would g(f) be??.......

We have..........g[f] = 2[f[ = 2[3x] = 6x........Note that all I did was to "stick" the function "f" - in this case, 3x - into the "box" in function "g" .......

So...what we're REALLY doing is, instead of putting a NUMBER into the "x's," we're putting a "function' into them!!!

Let's look at your problem, now. (finally!!!)

find [goh](x)
g(x)=x+3
h(x)=x^2+5x-7

OK.........let g(x) be written as [ ] + 3

Now......the notation (g o h)(x) just means the same thing as g(h).....Notice that the SECOND function listed - (h) - is being put into the FIRST one - (g)!!

Well.....let's do that now!!!

We have.........g(h) = [h] + 3 = [x^2 + 5x - 7] + 3 = x^2 + 5x - 4.......!!!!........ And that's your answer

Additionally. what if we wanted to know (h o g)(x) = h(g), instead??

We have h(g) = [g]^2 + 5[g] - 7 = [x + 3]^2 + 5[x + 3] - 7 = ( x^2 + 6x + 9) + (5x + 15) - 7 = x^2 + 11x + 17 !!!!!!........ Notice, I just "stuck" "g" into "h'

And that's it!!!

I hope you see it, now. It's kind of a tough concept to master, at first!!
18 mar. 2014
 #1
avatar+104899 
+3
Yours is an interesting question....and a little hard to think about!!

I'll give it a go and invite others to correct my logic flaws - of which I hope there aren't any.!!!

Let's consider a couple of scenarios;

Suppose I wanted to increase the speed of something twice. Then, this is just a 100% increase. Now, suppose that a process originally took 6 seconds and I wanted it to take just 3 seconds - i.e., twice as fast. Then, I could take 100% and covert it to a decimal, i.e., 1, and then add 1 to this which would give me 2. Now take the reciprocal of this (1/2) and multiply it by the original 6 seconds. So, (1/2)*6 = 3 seconds, and I've doubled the speed.

Well....so far, so good. Now suppose I wanted to triple the speed of some process. Following the above logic, this would be a 200% increase. So, convert 200% to a decimal, i.e., 2, and add 1 to this = 3. Now take the reciprocal of this = (1/3). So, if a process took 6 seconds and I wanted to triple the speed, I'd just multiply the (1/3) times 6 and get 2 seconds - triple the speed of what it originally was.

So....since you want a 55% increase in speed, let's convert this to a decimal (.55) and add 1 to it = 1.55. Now take the reciprocal of this (1/1.55) and multiply it by the time interval of the original process - 2 seconds,. So we have (1/1.55)*(2) = about 1.29 seconds. Thus, a process that originally took 2 seconds now takes 1.29 seconds and we've sped up the process by 55%.

Look this over. I believe it's what you asked for. (Feel free to point out any perceived errors!!)
18 mar. 2014