A Basic Function Problem with a Limit

[DMOC]

Homework Statement

For all real numbers x and y, let f be a function such that f(x+y)=f(x)+f(y)+2xy and such that the limit as h approaches zero is f(h)/h=7. \stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7

a) Find f(0) and justify your answer.

b) Use the definition of the derivative to find f ' (x).

c) Find f(x).

NO CALCULATOR USE ALLOWED!

Homework Equations

Limit as h->0 is (f(x+h)-f(x))/h

The Attempt at a Solution

To solve part a, I looked at the original equation. To find f(0), I just put in 0 for x and 0 for y.

Let x=y=0.

And thus, I got f(0)=0, so now part a should be cleared. Right?

Then I get to part b and I'm a little stuck. Find f ' (x) ? Should I be using the equation I put in the relevant equations section to help me out with this?

 

[aPhilosopher]

The definition of the derivative is the equation that you gave, so yes.

 

[DMOC]

OKay, so here goes my shot at part b (do you think part a is right?):

f ' (x) is ...


(f(x+7) - f(x)) / 7

Hm ... I'm getting a little lost here. I just substituted h for 7.

Also, is there a way to post these fractions using the latex reference?

 

[aPhilosopher]

you want to start with:
\stackrel{Lim}{h\rightarrow0}\frac{f(x + h) - f(x)}{h}


What can you say about f(x + h)?

to do the fractions, just use \frac{}{} where the numerator goes in the first bracket pair and the denominator goes in the second.

 

[DMOC]

you want to start with:
\stackrel{Lim}{h\rightarrow0}\frac{f(x + h) - f(x)}{h}


What can you say about f(x + h)?

to do the fractions, just use \frac{}{} where the numerator goes in the first bracket pair and the denominator goes in the second.

Thanks for letting me know about the fractions part.

For f(x+h), I was thinking that would just be equal to f(x) since h is approaching zero. However ...

\stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7

Which doesn't make sense anyway since that would be undefined as h approaches zero. Hm ...

 

[aPhilosopher]

You have the formula for f(x + [strike]y[/strike]h) in your original post ;)

 

[aPhilosopher]

Also, \stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7 according to your original post. You edited it but left the old version. Which one is it?

 

[DMOC]

Sorry, the limit as h goes to zero is definitely SEVEN.

So how about this. I was thinking that since I could do this ...

f(x+y) = f(x) + f(y) + 2xy
f(x+y) - f(x) = f(y) + 2xy

Then couldn't I say that since h is approaching zero, then it figures that:

x=y=0=h as well. I can then substitute the h's in for x or y?

Am I on the right track here?

 

[aPhilosopher]

well f(x+h) - f(x) = f(h) + 2xh is right but that's inside a limit and over h as h goes to zero.

now, as I'm sure you know, the sum of limits is the limit of sums. So do it like that. There is no y in this equation and you can't set x to a value because you're taking the limit for all x so it needs to remain a variable.

 

[DMOC]

I'm not sure what to do what you said, because we never discussed limits in my class yet.

 

[aPhilosopher]

Ok. I don't think there's a way to do this without limits as they're in the problem as stated so the rule you want is:
\lim_{h\to a}(f(h) + g(h)) = \lim_{h\to a} f(h) + \lim_{h\to a} g(a)

See if that helps.

 

[DMOC]

Ok. I don't think there's a way to do this without limits as they're in the problem as stated so the rule you want is:
\lim_{h\to a}(f(h) + g(h)) = \lim_{h\to a} f(h) + \lim_{h\to a} g(a)

See if that helps.

I think this problem requires knowledge of limits, although I never learned that last year (I think my teacher expected us to know from last year).

So for part b:

f ' (x) = \lim_{h\to 0} (f(x+h)/h) + \lim_{h\to 0} f(x)

Hopefully this is on the right track.

 

[aPhilosopher]

That's the idea but do you remember how to expand f(x + h)?

Just keep it in one limit until you have expanded f(x + h) and the f(x) cancels out like it did before. And remember that when a sum is on top of a fraction, the denominator goes with both sides of the sum if you split it up!

\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}

 

[DMOC]

f ' (x) = \lim_{h\to 0} (f(x+h)/h) + \lim_{h\to 0} f(x)

Revised version.

f ' (x) = \lim_{h\to 0} ((f(x+h)/h) + f(x/h))

 

[DMOC]

And, no I do not know how to expand f(x + h). I know this is easy once I understand the concept but I'm really new to it, sorry.

 

[aPhilosopher]

That's ok. Look at your original post and post #8 and #9 in this thread. You've already done it

 

[DMOC]

Really?

So do I substitute f(h) + 2xh for f(x+h) - f(x) now?

I'm still a bit confused, sorry.

 

[aPhilosopher]

No problem.

Try it and see what happens. Post your results!

 

[DMOC]

\stackrel{Lim}{h\rightarrow0}\frac{f(y) + 2xy}{h}

Do I put an equal sign after or before the limit part?

 

[aPhilosopher]

Ok, where is the y coming from?

 

[DMOC]

The y comes from ( f(y) + 2xy ) being equal to ( f(x+h) - f(x) ).

 

[aPhilosopher]

Ok, but you're taking the limit with respect to h so it starts of as f(x + h) - f(x). Why are you switching to y?

 

[DMOC]

Not sure, really, since a classmate of mine told me that I might have to do something like that. Yeah, clueless.

Should I set h equal to something to help me find the derivative of f ' (x) ?

 

[aPhilosopher]

No, you can't set anything equal to anything.

you have \lim_{f \to 0}\frac{f(h) + 2xh}{h} given the rule I gave you earlier for the sum of a limit, can you figure it out from there? you already know what \lim_{f \to 0}\frac{f(h)}{h} is. How does the value of 2x change when h changes? (assuming that x is fixed)

 

[DMOC]

\lim_{h \to 0}\frac{f(h)}{h} + \lim_{h \to 0}\frac{2xh}{h}

The second part would just cancel out since it's a straight up 2xh/h problem wher ethe h's cancel. 2x remains.

f ' (x) = 7 + 2x

Just wondering, but how did you replace f(x) with 2xh?

 

[aPhilosopher]

I didn't. We replaced it with 7 because that's what the problem specifies! The h's do cancel when taking limits. So now how do you get f?

 

[DMOC]

Where does the problem specify that f(x) = 7?

----------------

Okay, so f ' (x) = 7 + 2x seems to be the answer for part b.

Now for f(x), that equals:

f(x + y) - f(y) - 2xy

Am I supposed to put in 0 for y?

 

[aPhilosopher]

Sorry, i was being sloppy. I shouldn't do that. \lim_{h \to 0}\frac{f(h)}{h} = 7 is specified in the problem.

Do you know how to integrate?

We're not going to be setting anything to anything for the rest of the problem so quit thinking about it! ;) It was a good trick for part (a) but we don't need it anymore.

 

[DMOC]

I have no idea what integrating is ... our teacher expected us to know stuff from last year, but I took a different class from the rest of my classmates. :(

So any clues on how to find f(x)?

 

[aPhilosopher]

Well, do you know how to take derivatives?