Beginning Numerical Analysis question. Still calculus

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Homework Help Overview

The discussion revolves around demonstrating that the function f(x)=(x−2)sinxln(x+2) has a derivative f'(x)=0 at some point within the interval [-1,3]. Participants are exploring the application of calculus theorems and numerical analysis techniques to address this problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of Rolle's theorem and the mean value theorem, noting challenges with the conditions required for these theorems. Some suggest checking specific values within the interval, while others propose generating numerical data to analyze changes in monotonicity.

Discussion Status

There is an ongoing exploration of different methods to approach the problem. Some participants have suggested generating numerical values to identify monotonicity changes, while others are reconsidering the application of Rolle's theorem with a modified interval. The conversation reflects a mix of theoretical and numerical strategies without a clear consensus on the best approach.

Contextual Notes

Participants are navigating the constraints of the problem, including the need for specific conditions to apply certain theorems and the implications of numerical analysis in this context. There is also a discussion about the relevance of the interval [0,1] versus [0,2] in relation to applying Rolle's theorem.

jaqueh
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Homework Statement


Show f(x)=(x−2)sinxln(x+2) has f'(x)=0 somewhere on [-1,3]

The Attempt at a Solution


I tried using Rolle's theorem, but f(-1)≠f(3). Then I tried the mean value theorem, but didn't get 0 either.
 
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Check some obvious values in between -1 and 3. Maybe x=0 and x=2? You are giving up too easily.
 
jaqueh said:

Homework Statement


Show f(x)=(x−2)sinxln(x+2) has f'(x)=0 somewhere on [-1,3]

The Attempt at a Solution


I tried using Rolle's theorem, but f(-1)≠f(3). Then I tried the mean value theorem, but didn't get 0 either.

You can find f'(x) analytically, so maybe you should do that and then apply the intermediate value theorem.
 
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jaqueh said:

Homework Statement


Show f(x)=(x−2)sinxln(x+2) has f'(x)=0 somewhere on [-1,3]

The Attempt at a Solution


I tried using Rolle's theorem, but f(-1)≠f(3). Then I tried the mean value theorem, but didn't get 0 either.

Well if it's numerical analysis, why are you using those theorems? Suppose it was some function that you didn't even know what the form was, how could you show the derivative is zero somewhere in that interval? If it was mine, I'd generate "numerically" since this is numerical analysis, a set of equally spaced points in that interval and then compute the value of the function and then I'd inspect the list for the "monotonicity" change, that is, when the numbers are increasing then decreasing or vice-a-versa. However if there is an inflection point, the derivative could still be zero without this change in monotonicity. But we could at least rule out the former case this way. Anyway, lots of tough problems won't give you the luxury of using nice theorems and you'll have to muscle-through the data in this way or another.

I got an idea, why don't you write a short Mathematica program to generate these numbers then pick out the monotonicity change?
 
I got it, make [0,1] a subset of [-1,3] then i can use Rolle's theorem and then determine that there must be a c in [0,1] so there's got to be a c in [-1,3]
 
jaqueh said:
I got it, make [0,1] a subset of [-1,3] then i can use Rolle's theorem and then determine that there must be a c in [0,1] so there's got to be a c in [-1,3]

What does the interval [0,1] do for you?
 
LCKurtz said:
What does the interval [0,1] do for you?

sorry i meant for the interval to be [0,2]
then i get f(0)=f(2)=0 => rolle's
 
Last edited:
jaqueh said:
sorry i meant for the interval to be [0,2]
then i get f(0)=f(2)=0 => rolle's

That's better. :smile:
 
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