I have to prove x^3 is differentiable at x=4

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Discussion Overview

The discussion revolves around proving that the function \( f(x) = x^3 \) is differentiable at \( x = 4 \) using the formal definition of differentiability. Participants explore the steps involved in the proof and the level of rigor required for different calculus courses.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes that to show differentiability at \( x = 4 \), it suffices to demonstrate that \( f'(4) = \lim_{x \to 4} \frac{f(x) - f(4)}{x - 4} \) exists.
  • Another participant suggests that the proof is acceptable at an elementary calculus level but may require more formalism in advanced calculus contexts.
  • A different participant humorously implies that the proof would be deemed unacceptable in a Real Analysis course.
  • One participant emphasizes the need for justifying each step in the proof, mentioning that continuity preserves convergence and that the limit does not require the function to be defined at the point itself.
  • Another participant reiterates the previous point, adding that the proofs for advanced calculus and calculus are fundamentally similar, with the addition of more rigorous epsilon-delta arguments.
  • One participant discusses the equivalence of functional limit definitions and sequence limit definitions, arguing that due to continuity, the limit of a polynomial at a point is simply the polynomial evaluated at that point, suggesting that epsilon-delta arguments may be unnecessary.
  • A participant repeats their initial proof attempt, reinforcing that "differentiable at \( x = a \)" means that the limit exists.

Areas of Agreement / Disagreement

Participants express varying opinions on the rigor required for the proof, with some suggesting that additional justification is necessary while others believe the initial proof is sufficient. There is no consensus on the exact requirements for different levels of calculus courses.

Contextual Notes

Some participants note that the proof may need to be adjusted based on the course level, indicating a potential dependence on the definitions and expectations set within different educational contexts.

Caeder
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I have to prove x^3 is differentiable at x=4 using the definition of what it means for something to be differentiable.

So I was wondering if I just have to show that f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4} exists, where f(x) = x^3.

So...

f'(4) = \lim_{x \to 4} \frac {x^3 - 64}{x - 4}

= \lim_{x \to 4} \frac{(x-4)(x^2 + 4x + 16)}{x-4}

= \lim_{x \to 4} x^2 + 4x + 16.

So the limit (derivative) is 48...does this prove it's dif'able at x=4?
 
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The proof is acceptable at an elementary calculus level. If you are taking advanced calculus, you would need a little more formalism, but the idea is the same.
 
Ha, then I bet for a Real Analysis course this proof would be unacceptable :-p
 
actually all you need to do is add a few lines of text justifying each step. Use facts like continuity preserves convergence and the functional limit at a point x does not even require the function to be defined at the point, just near it.

That aside the proofs for adcalc/analysis and calculus are identical
 
SiddharthM said:
actually all you need to do is add a few lines of text justifying each step. Use facts like continuity preserves convergence and the functional limit at a point x does not even require the function to be defined at the point, just near it.

That aside the proofs for adcalc/analysis and calculus are identical

With a few more epsilons and deltas in the mix ;)
 
JonF said:
With a few more epsilons and deltas in the mix ;)

once you prove that functional limit definition is equivalent to the sequence limit definition (this is analogous to continuity and sequential continuity the difference being you don't consider the point which you are approaching, in fact f doesn't have to be defined at that point for the limit to exist) you see that because of continuity

the limit of a polynomial at x is the polynomial evaluated at x.

so that any epsilon-delta argument is unnecessary.
 
Caeder said:
I have to prove x^3 is differentiable at x=4 using the definition of what it means for something to be differentiable.

So I was wondering if I just have to show that f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4} exists, where f(x) = x^3.

So...

f'(4) = \lim_{x \to 4} \frac {x^3 - 64}{x - 4}

= \lim_{x \to 4} \frac{(x-4)(x^2 + 4x + 16)}{x-4}

= \lim_{x \to 4} x^2 + 4x + 16.

So the limit (derivative) is 48...does this prove it's dif'able at x=4?
Because "differentiable at x= a" means precisely that that limit exists!
 

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