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f_ball_master · Oct 27, 2009, 06:52 PM

If g(x)=2x^2+3 find:
g(2+h)-g(2)/h

Nhatkiem · Oct 27, 2009, 08:03 PM

Originally Posted by f_ball_master

If g(x)=2x^2+3 find:
g(2+h)-g(2)/h

I believe what your posting is the difference quotient.

if we have $g(x)=x^2+x$

and the difference quotient asks for $\frac{g(x+h)-g(x)}{h}$

Then the way I would go about it is inputing "x+h" and "x" into my function g(x). Like so:

$\frac{g(x+h)-g(x)}{h}=\frac{[(x+h)^2+(x+h)]-[(x)^2+(x)]}{h}$

I'm using multiple parenthesis here in the hopes that you more easily see what I'm plugging in, and where I'm plugging it into.

So we have, for this example:

$\frac{g(x+h)-g(x)}{h}=\frac{[(x+h)^2+(x+h)]-[(x)^2+(x)]}{h}$
simplify (x+h)^2
$\frac{g(x+h)-g(x)}{h}=\frac{[x^2+2xh+h^2+x+h]-[(x)^2+(x)]}{h}$
simplified the expression further by removing some un-needed parenthesis
$\frac{g(x+h)-g(x)}{h}=\frac{[x^2+2xh+h^2+x+h-x^2-x]}{h}$
distributed the negative
$\frac{g(x+h)-g(x)}{h}=\frac{[x^2+2xh+h^2+x+h-x^2-x]}{h}$
combined like terms
$\frac{g(x+h)-g(x)}{h}=\frac{[2xh+h^2+h]}{h}$
divided through by h each term
$\frac{g(x+h)-g(x)}{h}=2x+h+1$

Now this is just an example, but it shows all the needed steps to do this, hopefully it helps you do the problem. Post back if you need any help.

morgaine300 · Oct 27, 2009, 09:16 PM

This is how I do these. Forget that h in the denominator for a minute. You can throw that back in once you get the numerator set up, which is the hard part.

When you look at
$g(2+h)-g(2)$
take away what is in parenthesis for a minute. Forget the 2+h and 2 are there. Think of it as just:
$g-g$ .

Now what is g?

$2x^2+3$

Substitute that equation in each g. Yes, it's the same equation. Don't worry about that. After all, it is saying g minus g. So plug in that g equation:

$[2x^2+3]-[2x^2+3]$

So now you have g minus g. I have found for many people that it helps if they set off the variable in parenthesis, just for the sake of the visual on the next part:

$[2(x)^2+3]-[2(x)^2+3]$

Once you have this done, you're set to insert the amount they've given for x. Don't try to look at that g(2+h) and make it complicated. Keep in mind that a function equation is basically saying that given some particular x... and we could insert a given x and solve it.

Like a simple one like this:

$f(x)=2+x$
is saying that for any given x, do 2+x. And we can insert any x that we like. And it will be put into those parenthesis, right? So whatever is in parenthesis will be inserted wherever the x is. So if I insert 3 I get:

$f(3)=2+3 = 5$

Notice I'm inserting whatever is in those parenthesis into wherever the x is.

So now you go back to what you have been given. The first g(x) says g(2+h). Don't try to make anything of the h. Just think about the fact that (2+h) is what is inside that parenthesis. So that is exactly what you will insert into your first g(x), right where the x is. That is why separating off the x in parenthesis can help, to make sure you get it into the right place -- that is, don't get confused and think the "+h" part has anything to do with the "+ 3" part. It doesn't. Insert that whole (2+h) right where the x is.

Then for the second g(x) that's easier. That one says x is going to be 2, so just insert 2 for x in the second one.

Don't forget to put the h back into the denominator. :)

Then solve. Be sure to use lots of []'s or {}'s to keep everything straight.