(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that:

Any function [itex] f [/itex] such that [itex] f(x)=x^3 [/itex] for any [itex] x \in [/itex]Ris differentiable.

2. Relevant equations

Skip.

3. The attempt at a solution

Okay! So, to conclude, it must be shown that, for any [itex] a [/itex] in the domain of [itex] f [/itex],

[itex] \displaystyle \exists \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} [/itex].

Quickly,

[itex] \displaystyle \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} [/itex]

[itex] {} = \displaystyle \lim_{h \rightarrow 0} \left( 3a^2 + 3ah + h^2 \right) [/itex].

Now, the latter limit must be proved (via delta-epsilon method) to equal [itex] 3a^2 [/itex]. (This is the part I am having most trouble with, where delta-epsilon proofs are still sort of a mystery to me.)

Suppose [itex] \varepsilon > 0 [/itex] is given. So, we should be able to show that if [itex] 0 < \left| h \right| < 0 [/itex] for some [itex] \delta > 0 [/itex], then we have [itex] \left| \left( 3a^2 + 3ah + h^2 \right) - 3a^2 \right| = \left| h \right| \cdot \left| 3a + h \right| < \varepsilon [/itex]. Well, we could choose [itex] \delta = \varepsilon / \left| 3a + h \right| [/itex], however we must limit the size of [itex] \left| 3a + h \right| [/itex] in terms of our [itex] 0 < \left| h \right| < 0 [/itex] statement (why must we do this?). With that said, suppose [itex] \delta = 1 [/itex]. Then, [itex] \left| h \right| < 1 [/itex]. Yet, this means [itex] \left| 3a + h \right| \leq \left| 3a \right| + \left| h \right| < 3 \left| a \right| + 1 [/itex]. Therefore, the claim (hopefully!) follows if we choose [itex] \delta = \min \left\{ 1, \varepsilon / \left( 3 \left| a \right| + 1 \right) \right\} [/itex].

QED

(Thanks for any help and or advice!)

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# Prove that Any Cubed Function is Differentiable (delta-epsilon method)

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