# Math Help - f(x) = x^3

1. ## f(x) = x^3

Let $f(x) = x^3$.

1.) Prove (using the $\epsilon-\delta$ definition) that at $x = 2, f$ is continuous on $\mathbb{R}$.

2.) Prove (using the def. of differentiability) that at $x = 4, f$ is differentiable.

Well, I don't know if this is helpful, but I know $f$ is not uniformly continuous on $\mathbb{R}$. There is a similar example in my book proving that x^2 is continuous on $\mathbb{R}$. They do:

Given $c \in \mathbb{R}$:

$|f(x) - f(c)| = |x^2 - c^2| = |x-c||x+c|$

Now they find an upperbound on $|x+c|$ by makings ure $\delta$ does not exceed 1. Apparently this insures all values of $x$ will fall in the interval $(c-1,c+1)$.

I don't entirely get that ste.

Then, they get:

$|x + c| \leq |x| + |c| \leq (|c|+1) + |c| = 2|c| + 1$

They let $\epsilon > 0$ be given, then choose $\delta = min\left(1,\frac{\epsilon}{2|c|+1}\right)$, then $|x - c| < \delta$

$\implies |f(x) - f(c)| = |x-c||x+c|< \frac{\epsilon}{2|c|+1}(2|c|+1) = \epsilon$.

I'm assuming it's similar for $x^3$?

2. Idea:

Can I use the Algebraic Continuity Theorem and show $f(x) = x$ is continuous, and therefore $f(x) = x^3$ will be continuous too? Since we can take f(x)*f(x)*f(x) = g(x), and by ACT, g(x) should be cont too.

If so, how does this proof look:

Let $\epsilon > 0$ be given. Further, choose $\delta = \epsilon$. Then, for $x,y \in \mathbb{R}, |x - y| < \delta \implies |f(x) - f(y)| = |x - y| < \epsilon$

3. see here

4. Oh nice! So, I can't make $\delta = \epsilon$ like my proof above?

5. Originally Posted by seerTneerGevoLI
Oh nice! So, I can't make $\delta = \epsilon$ like my proof above?
correct. use the structure of the proof given there

6. Originally Posted by seerTneerGevoLI
2.) Prove (using the def. of differentiability) that at $x = 4, f$ is differentiable.
How rigorous was your definition of differentiability

if not very rigorous, you can just show that the derivative limit has a numerical value using the limit definition of the derivative. that is, find

$\lim_{h \to 0} \frac {f(4 + h) - f(4)}{h}$

7. Let $f : A \rightarrow \mathbb{R}$ be a func. that's defined on an interval A. Given $c \in A$, the derivative of f at c is defined by:

$f'(c) = \lim_{x\rightarrow c}{\frac{f(x) - f(c)}{x-c}}$

provided that the limit exists.

$f$ is diff'able on $A$ if $f'$ exists for all points $c \in A$.

8. Originally Posted by seerTneerGevoLI
Let $f : A \rightarrow \mathbb{R}$ be a func. that's defined on an interval A. Given $c \in A$, the derivative of f at c is defined by:

$f'(c) = \lim_{x\rightarrow c}{\frac{f(x) - f(c)}{x-c}}$

provided that the limit exists.

$f$ is diff'able on $A$ if $f'$ exists for all points $c \in A$.
fine, so all you need to do is show that $f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4}$ exists, where $f(x) = x^3$ of course.

Tip: You can use "\to" instead of "\rightarrow" for the $\to$ symbol in your limit

9. For the second one:

$f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4}$

So...

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

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

$= \lim_{x \to 4} x^2 + 4x + 16$ ONLY IF $x \neq 4$.

So I don't see how to prove this is dif. at 4.

Suppose I could do this. The limit then is 48. Is this really a proof showing it's dif'able?