Someone helps me please. What is the relationship between Ax =0 and Ax=b
I think b = 0
okay
why did u ask
Let r>0 \(f(x)=\begin{cases} x^rsin(1/x) ~~~~if x>0\\0~~~if x\leq 0\end{cases}\) how to prove f(x) differentiable?
@freckles
Well we need first establish that f is continuous at x=0 and then we need to check out that whole smoothness thing.
\[f(0)=0 \text{ is easy \to see now we just need \to \check if that limit thing is 0 } \\ \lim_{x \rightarrow 0^+} x^r \sin(\frac{1}{x}) =\lim_{x \rightarrow 0^-}0\] so you know that left limit at x=0 is also obviously 0 we just to make sure the other side is also 0 then we can move on to that smoothness requirement
\[-1 \le \sin(\frac{1}{x}) \le 1 \] I suggest using squeeze theorem
Still need me, or you got it?
how? when left side is -1 and right side is 1
multiply both sides by x^r
all sides*
I need a lot of help. Appreciate any help
\[-x^r \le x^r \sin(\frac{1}{x}) \le x^r \]
now we know r>0 so you know both -x^r and x^r go to what as x goes to zero?
not quite how squeeze is applied there, but yes, you need to use the periodic nature of sine. Then squeeze if you have already proved it in class
0
by the way the direction of the inequality isn't effected by our multiplication because x>0 since we are only looking at x goes to infinity so you just did the first part you showed that the function is continuous at x=0
so for any value of r, we have f'(0) exist, right?
now we need to check the smoothness at x=0 (I thought we were looking at r>0)
ok, that part is ok. Now another question: :) for which value of r, f'(x) continuous at 0?
well you know we haven't finished your first problem right?
anyways...for the second question you have here you need to find f'(x)
do that by finding it in pieces first
like find the derivative of both left and right sides of the function at x=0
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