Finding f'(0) with L'Hopital's Rule and a Piecewise Function

In summary, the student tried to find a differentiable function at 0 by using L'Hopital's rule multiple times, but was not able to find a valid equation. However, if you are only interested in the value of f'(0), any choice of g(0) will do.
  • #1
FsLiu
21
0

Homework Statement



Find f'(0) given a piecewise function defined as

f(x) =
{g(x)/x2, x≠0
{0, x=0

where g(x) is a function satisfying g(0)=g'(0)=g''(0) and g'''(0)=14


Homework Equations


none.

The Attempt at a Solution



So far, I've reasoned that for f to be differentiable at 0, limit as x approaches 0 of g(x)/x2 is zero, and that leads to L'Hopital's rule (?). However, I get stuck after 2 derivations with 0/2 where the rule no longer applies, and I'm unable to use the fact that g'''(x)=14.

Any help would be much appreciated :smile:
 
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  • #2
I'm assuming that g(0)=0. You forgot to mention that.

It is indeed true that after 2 derivations you get 0/2. This is correct. This means that the limit is 0. Thus you have found that

[tex]\lim_{x\rightarrow 0}{\frac{g(x)}{x^2}}=0[/tex]

But this only means that f is continuous at 0. You need to calculate another limit to find that f is differentiable at 0.
 
  • #3
micromass said:
I'm assuming that g(0)=0. You forgot to mention that.

It is indeed true that after 2 derivations you get 0/2. This is correct. This means that the limit is 0. Thus you have found that

[tex]\lim_{x\rightarrow 0}{\frac{g(x)}{x^2}}=0[/tex]

But this only means that f is continuous at 0. You need to calculate another limit to find that f is differentiable at 0.

Yeah, forgot to mention that sorry.

I actually took a different approach to the question while I was thinking about it in class. I figured that using L'Hopital's rule multiple times only seems to prove that the function is continuous.

I integrated g'''(x) multiple times to get an equation g(x)=7x3/3, making f(x)=7x/3 and then f'(0)=7/3. Is this valid? working backwards, everything seems to be consistent, only the piecewise part is just making me feel somewhat uneasy about doing it this way.
 
  • #4
But you only know the value of g in 0. For all you know, the function will be something very different...
 
  • #5
micromass said:
But you only know the value of g in 0. For all you know, the function will be something very different...

Wait, but g'''(0)=14

integrate that, g''(x)=14x+C, C=0 since g''(0)=0 through substitution

integrate, g'(x)=7x2+C, again C=0 through substitution

integrate, g(x)=7x3/3+C, C=0 through substitution

then plug that g(x) into the piecewise function to get f(x)=7x/3

Seems to work, doesn't it?
 
  • #6
I'll give you a counterexample. The function

[tex]g(x)=\frac{14}{6}x^3e^x[/tex]

is also a good choice for g.
 
  • #7
micromass said:
I'll give you a counterexample. The function

[tex]g(x)=\frac{14}{6}x^3e^x[/tex]

is also a good choice for g.

ahh, now I'm confused as f**, what did I do wrong in the integrals?

Also, the counterexample you gave does yield the same answer to f'(0)=7/3, and just out of curiosity, how did you come up with that counterexample so quickly, I'm quite baffled, are you a mathematician or prof or something?
 
  • #8
FsLiu said:
Wait, but g'''(0)=14

integrate that, g''(x)=14x+C
No, you can't say this.

If it was given that g'''(x) = 14, then you could say that g''(x) = 14x + C, but you don't have any information about g''(x), only its value when x = 0.
FsLiu said:
, C=0 since g''(0)=0 through substitution

integrate, g'(x)=7x2+C, again C=0 through substitution

integrate, g(x)=7x3/3+C, C=0 through substitution

then plug that g(x) into the piecewise function to get f(x)=7x/3

Seems to work, doesn't it?
 
  • #9
Mark44 said:
No, you can't say this.

If it was given that g'''(x) = 14, then you could say that g''(x) = 14x + C, but you don't have any information about g''(x), only its value when x = 0.

Wait but the original question stated that g(0)=g'(0)=g''(0)=0, sorry I forgot to edit that in.

I asked a second year student about it though, and he thinks it'd suffice because I found one general solution to the problem, and the question is only asking for f'(0) which I found to be 7/3.
 
  • #10
No, it doesn't suffice to say that g(x)=7x³/3+C, because it is simply not true. g could be many other things, includin that example I gave you...

You'll need to check that f(x) is differentiable by checking the definition. What is the definition for "f is differentiable"?
 
  • #11
It may not suffice, but what is true, assuming g(x)=7x3/3, is g'(0)=g''(0)=0 and g'''(0)=14.

And the question asks for the value of f'(0), so any possible equation of g(x) would suffice.

Thanks for all the input guys, but I'm going conclude that question there and ask the TA's/profs about it tomorrow. Brain hurts from this.
 
  • #12
Yes, if you're interested in just the value of f'(0), then any choice of g(0) should do.
However, how do you know that every choice of g(x) yields the same value for f'(0)??
 

1. What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical principle used to evaluate limits of indeterminate forms, commonly used in calculus. It states that for a function f(x) and g(x) that both approach 0 or infinity as x approaches a certain value, the limit of their ratio can be found by taking the derivative of both functions and evaluating the resulting limit.

2. When is L'Hopital's Rule applicable?

L'Hopital's Rule is applicable when evaluating limits of the form 0/0 or infinity/infinity. It can also be used for limits of the form 0*infinity or infinity - infinity, by manipulating the expressions to fit the 0/0 or infinity/infinity form.

3. How do I use L'Hopital's Rule?

To use L'Hopital's Rule, first identify the limit of the given function. Then, take the derivative of both the numerator and denominator of the function. If the resulting limit still has an indeterminate form, repeat the process until a definite value is obtained or the limit is proven to be infinite.

4. Can L'Hopital's Rule be used for all types of functions?

No, L'Hopital's Rule is only applicable for evaluating limits of indeterminate forms. It cannot be used for evaluating limits of functions that approach a finite value as x approaches a certain point.

5. Are there any limitations to using L'Hopital's Rule?

Yes, L'Hopital's Rule is limited to functions that are differentiable. It cannot be used for functions that are not continuous or have points of discontinuity. Additionally, it may not always provide a definite value for the limit, and other methods may need to be used to evaluate the limit.

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