I want to understand how to use the general equation [f(x) - f(x+h)]/h

[esanoussi]

I am in a 12th grade Physics class and in the beginning of the year, we learned the calculus based equation: [f(x+h) - f(x)]/h. I understand that I am trying to simplify/solve for the answer when H goes to 0 (h -> 0), and I understand that I must substitute f(x) into the given fields f(x) and f(x+h). However, I can never manage to completely simplify the problem.

Here is a sample problem we were given in Physics class earlier this year:

f(x) = x²+1/x

Solve for: [f(x+h) - f(x)]/h


Can someone please help me understand how to do these sort of problems? Is there any kind of technique that can help me solve them? I need to know ASAP, as I am struggling in class.

 

[arildno]

Well, f(x+h) then replaces "x" in the general formula woth "x+h" at all instances:
f(x+h)=(x+h)^{2}+\frac{1}{x+h}
The difference is then:
f(x+h)-f(x)=x^{2}+2xh+h^{2}+\frac{1}{x+h}-(x^{2}+\frac{1}{x})=2xh+h^{2}+\frac{x}{x(x+h)}-\frac{(x+h)}{x(x+h}}=h*(2x+h-\frac{1}{x^{2}+xh})
Whereby dividing by "h" yields, for every non-zero "h":
\frac{f(x+h)-f(x)}{h}=2x+h-\frac{1}{x^{2}+xh}

As h tends to 0, we will get:
f'(x)=2x-\frac{1}{x^{2}}

 

[esanoussi]

arildno,

Thank you for helping me solve the problem. While I understand how you arrived at the answer, as you probably know, every physics problem is different. Is there any technique I can use, or any tip you can provide me when solving future problems and more complex problems using the same formula? Any advice you can give would be much appreciated.

 

[arildno]

1. "every physics problem is different. "
Not so.

For example, now you should know how to calculate the difference expression for ANY function f.

Try a few more here on PF, and we will help you if you get stuck in the middle.

Now is the time to hone your skills on THESE types of problems.

 

[esanoussi]

So every problem isn't different? Oh. We were taught in Physics class that every problem is.

Thanks so much for your help. I will take your advice and check out other problems on the forum.

 

[arildno]

It depends on what you mean by "different".

The same general technique can be used to solve a lot of similar problems.

 

[emyt]

So every problem isn't different? Oh. We were taught in Physics class that every problem is.

Thanks so much for your help. I will take your advice and check out other problems on the forum.

this is a general method in itself, the idea is to algebraically manipulate the equation until you can get something that isn't a division by zero

think of "excluding" the case where h is "literally" zero, since you want to consider the behaviour of the function as h APPROACHES zero. So, if you see something that doesn't work because "it might be zero", then ignore that it might be zero and do it anyway