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