# Thread: I have no idea what this is called or how to solve

1. ## I have no idea what this is called or how to solve

Using the given function, find f(x+h) - f(x) / h

A.) f(x) = 3x-7

B.) F(X)=X^2 +3X -2

I would show work, but I have no idea what to do... I know the answers I just really want to know how to solve.

Thanks

2. Okay. So the problem is asking you to substitute various expressions for f(x) into the formula

$\displaystyle \frac{f(x+h)-f(x)}{h}$

So all you do is take the expression and plug it in where f(x) is. For example, for the first one the expression is 3x-7. So we want to first replace any f(x) we find with 3x-7:

$\displaystyle \frac{f(x+h)-(3x-7)}{h}$

Second, we see that f(x+h) is also a function, even though it doesn't look exactly like f(x). What we have to do to figure out f(x+h) is do the exact same thing we did with the regular f(x) - we substituted 3x-7 - except instead of just plain x, we do it with x+h. So:

$\displaystyle f(x+h) = 3(x+h)-7$

Now, since we've figured out f(x+h), we can substitute the value we got back into the original equation. Now we have:

$\displaystyle \frac{3(x+h)-7 -(3x-7)}{h}$

When we multiply this out and simplify algebraically, we get

$\displaystyle \frac{3h}{h} = 3$

What this is is really a beginning calculus problem; it's the formula for finding a derivative using limits. So you might get better answers trying the calculus board. Can you solve the second one on your own? If not, and if I didn't explain in enough detail, just ask and I'll try to help you out some more.

3. Originally Posted by dto

Using the given function, find f(x+h) - f(x) / h

A.) f(x) = 3x-7

B.) F(X)=X^2 +3X -2

I would show work, but I have no idea what to do... I know the answers I just really want to know how to solve.

Thanks
It's a prelude to when you will be studying calculus (finding the derivative of the function f(x) from first principles).

4. Awesome Awsome! Thanks alot, cleared it all up.

So would the answer for part B, be 2x+h+3?

5. No problem! I'm glad you understood, cause I wasn't entirely sure I made sense. And yes, 2x+h+3 is totally right. Nice work!

It may be interesting to note that in most calculus problems, instead of the exact value of

$\displaystyle \frac{f(x+h)-f(x)}{h}$,

they would probably ask you to find the limit as h tends to zero:

$\displaystyle \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$

In this case, to find the limit you would just calculate the value of the expression like before, and then you would evaluate it when h=0. So,

$\displaystyle \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$ when $\displaystyle f(x) = x^2+3x-2$

is simply

$\displaystyle \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} = 2x+h+3$

and, just plugging in 0 for h, we get

$\displaystyle \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} = 2x+3$