Question

Let \(f(x) =\begin{cases}x^rsin(1/x)~~~if~~~x>0\\0~~~if~~~x\leq 0\end{cases}\) where \(r\in \mathbb R\), \(r>0\)
For which values of r is (f(x) everywhere differentiable?
For these values of r, find f'(x)
Please, help

Answer 1

@perl

Answer 2

This is what I got so far:
Suppose f(x) is everywhere differentiable,
if x>0, \(f(x) = x^r sin(1/x) \rightarrow f'(x) = rx^{r-1}sin(1/x)-x^{r-2}cos(1/x)\)
if \(x\leq 0\), \(f'(x)=lim_{x\rightarrow 0}\dfrac{f(x)-f(0)}{x-0}=lim_{x\rightarrow 0}x^{r-1}sin(1/x) =0\)
Hence the limit of \(0^+\)= limit of \(0^-\)
that is \(rx^{r-1}sin(1/x)-x^{r-2}cos(1/x)=0\)

Answer 3

So that
\(tan(1/x) =\dfrac{1}{xr}\)
And..... stuck.

Answer 4

(I don't know if this qualifies as a rigorous approach, so tread carefully.) For
\[f(x)=\begin{cases}
x^r\sin\dfrac{1}{x}&\text{for }x\gt0\\\\
0&\text{for }x\le0
\end{cases}\]
for positive real \(r\), you have the piecewise derivative
\[f'(x)=\begin{cases}
rx^{r-1}\sin\dfrac{1}{x}-x^{r-2}\cos\dfrac{1}{x}&\text{for }x\gt0\\\\
?&\text{for }x=0\\\\
0&\text{for }x\lt0
\end{cases}\]
The left-hand limit is clearly \(0\).

As for the right-hand limit, it only exists and is equal to \(0\) provided that \(r>2\). Suppose \(r=2\). Then
\[\lim_{x\to0^+}\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right)=0\pm1\]
which indicates non-existence.

Answer 5

I don't get how you are down to r >2.

Answer 6

You know that \(\lim\limits_{x\to0}\sin\dfrac{1}{x}\) doesn't exist, correct? The same goes for \(\cos\dfrac{1}{x}\). If you multiply a factor of \(x\), the oscillating nature of \(\sin\dfrac{1}{x}\)/\(\cos\dfrac{1}{x}\), the behavior of \(x\) will dominate the expression, so as \(x\to0\), so will \(x\sin\dfrac{1}{x}\) and \(x\cos\dfrac{1}{x}\).

In the left-hand piece of \(f'(x)\), you have a factor of \(x^{r-2}\) in front of \(\cos\dfrac{1}{x}\). When \(r>2\), the exponent is positive, which means as \(x\to0\), you have \(x^{r-2}\to0\) as well. Hence \(x^{r-2}\cos\dfrac{1}{x}\to0\) only if \(r>2\).

Otherwise, if \(r=2\), you have \(\cos\dfrac{1}{x}\to\pm1\) (so the limit does not exist); if \(r<2\), you have a negative exponent on \(x\), which means \(\dfrac{1}{x^{\text{power}}}\cos\dfrac{1}{x}\to\pm\infty\) (which is not \(0\)).

Answer 7

\(f'(x) = rx^{r-1}sin(1/x)-x^{r-2}cos(1/x) =0\)
We don't calculate limit again, right? just solve for r only, right?
How can we break it into limit?

Answer 8

You're given a function \(f:\mathbb{R}\to\mathbb{R}\) of the form
\[f(t)=\begin{cases}g(t)&\text{for }t\in X\\h(t)&\text{for }t\not\in X\end{cases}\]
where \(X\) is the interval \((0,\infty)\subset\mathbb{R}\).

Its derivative has the form
\[f'(t)=\begin{cases}g(t)&\text{for }t\in X\\h(t)&\text{for }t\not\in X\cup\{0\}\end{cases}\]
Note the latter case for \(h(t)\). We don't know for sure if \(f'(0)\) is defined because we don't know if \(f'(t)\) is continuous at \(t=0\). For this reason, we must compute the limits and establish equality to make a conclusion about the continuity (and hence existence) of the derivative at \(t=0\). We must satisfy
\[\begin{align*}\lim_{t\to0^-}f(t)&=\lim_{t\to0^+}f(t)\\
\lim_{t\to0}g(t)&\stackrel{?}{=}\lim_{t\to0}h(t)\end{align*}\]
If the one-sided limits are equal, then the two-sided limit exists. If the two-sided limit exists, then \(f'(t)\) is continuous at \(t=0\) provided that \(f'(0)=\lim\limits_{t\to0}f'(t)\). For this to be true for your given function, the limit must be \(0\).

The desired limit can only be \(0\) when \(r>2\). So when \(r>2\), the derivative at \(0\) must be \(f'(0)=0\).

Answer 9

@SithsAndGiggles I lost the net so that I couldn't answer you. I am sorry about that. :)

Answer 10

I got it. Thank you very much. :)