View Full Version : Functions

06-19-2011, 10:28 AM
I want to apologize up front if I do not write the question the right way for you.

Question: Given the functions g(x)= square root of x+7 and h(x)= 1/(2x-3) find g(h(x)) and h(g(x)).

I have already found the first one. The second one is confusing me.

So far I have h(g(x))= 1/(2square root of x+7) -3 *the minus 3 is in the denominator next to the square root*

I am not sure if I can take it any further?

06-19-2011, 11:48 AM
You posted this question before. viewtopic.php?f=9&t=47093 (http://www.freemathhelp.com/forum/viewtopic.php?f=9&t=47093)

Did you read my long answer? Tell me where my long answer left you stranded and we can take it from there. Sorry that my original explanation was not clear enough.

06-19-2011, 12:04 PM
Oh its okay. It is just really hard for me to understand math.

But on your long answer I am not sure where you got "^(1/2) from.

06-19-2011, 12:10 PM

x^(1/2) is the square root of x.

The ^ symbol (shift + 6) means exponentiation.

By the law of exponents [x^(1/2)] * [x^(1/2)] = x^[(1/2) + (1/2)] = x^1 = x.

Now let's try doing your problem using the method of substitution that I showed you in my long answer.

It is a slow method, but it makes things very clear. Just give me the first few lines and I'll tell you how you are doing OK?

PS When you do not understand an answer, just ask for clarification. No one around here bites though a few may bark from time to time.

06-19-2011, 12:20 PM
=1/((square-root x+7)-3)

This is as far as I can get. I want to say that I am suppose to get rid of the square-root in the denominator.

06-19-2011, 12:53 PM
To handle ANY problem at all dealing with composition of functions, such as g(h(x)) or h(g(x)), you can always substitute variables. It is a slow method, but it is sure and clear. g(h(x)) and h(g(x) are compositions of functions.

A function is a machine that spits out a number when you give it a number. That is all a function is.

Let u = g(x) = (x + 7)^(1/2).
Let v = h(x) = (1/(2x - 3)).
Let y = h(g(x)).

You work from the inside out. You feed x into the g machine. You feed the output of the g machine into the h machine.

g(x) = u.
So, h(g(x)) = h(u).
So what is h(u) based on the definitions given above?

06-19-2011, 01:20 PM
h(u)= 1/(2(x+7)^(1/2)-3)?

06-19-2011, 01:34 PM
h(u)= 1/(2(x+7)^(1/2)-3)?
Yes but you left out the step that makes it easy.

u = g(x) = (x + 7)^(1/2). If you say x = 9, (9 + 7)^(1/2) = 16^(1/2) = 4. You substitute 9 into the formula.

h(x) = 1 / (2x - 3).

So, h(u) = 1 / (2u - 3). You substitute u into the formula, just as you did with 9 above.

Clear so far?

So, h(g(x)) = h(u) = 1 / (2u - 3).

BUT u = (x + 7)^(1/2). You substitute that back into the formula above.
So h(g(x)) = 1 / (2((x + 7)^(1/2)) - 3), which is your answer. WELL DONE.

Do you see now how substitution (starting from the inside and working out) can help with composition of functions?

06-19-2011, 01:52 PM
I believe I understand now. I just have to slow it down.

06-19-2011, 01:57 PM
I believe I understand now. I just have to slow it down.
Yes, exactly.

There is frequently a fast way and a slow way. The slow way will prevent you from making most mistakes and will make clear WHY you are doing what you are doing.

After a while, it becomes automatic, and you can do it the fast way.

That start slow to understand and then go fast once you really have it is a good technique in all sorts of math.

06-19-2011, 02:35 PM

One more question...In that problem is it okay to leave the square root in the denominator?

06-19-2011, 02:44 PM

One more question...In that problem is it okay to leave the square root in the denominator?
Well, I suspect that is more a question of what formats, if any, are required by your teacher. Leaving the square root in the denominator is not wrong mathematically, and getting rid of it is going to be a mess if it is even possible.

06-19-2011, 03:01 PM
Oh okay. I was just wondering cause I know in the past we had some problems where a square root being in the denominator was not okay and you had to use the conjugate.