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$$ f_a(x) = \left\{ \begin{array}{ll} x^a & \quad x > 0 \\ 0 & \quad x \leq 0 \end{array} \right. $$

a.) For which values of $a$ is $f$ continuous at zero?

b.) For which values of $a$ is $f$ differentiable at zero? In this case, is the derivative function continuous?

Isn't $f$ always continuous? I'm somewhat confused how to approach this.

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    $\begingroup$ Consider $a=0$. Then is $f$ continuous? $\endgroup$ – MathQED May 10 at 22:15
  • $\begingroup$ No, then I would assume that is the only time it isn't. $\endgroup$ – mcs22 May 10 at 22:25
  • $\begingroup$ Are you sure? What if $a<0$? For example, $a=-1$? $\endgroup$ – MathQED May 10 at 22:27
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Why $f$ is not continuous at $0$ for some $a$:
Suppose $a=0$. Then $f(0)$ is undefined hence $f$ is not continuous at $0$.

(a) If $f$ is continuous at $0$, $f$ is defined on $0$ and $$f(0+)=f(0-)=f(0)$$ It can be shown tht $f(0-)=0$. However $f$ is not defined at $0$ for all $a\le0$. For $a\gt0$, $f(0)=0$. Therefore for $f$ to be continuous, $a\gt0$ and $f(0+)=0$.
We claim that for all $a\gt0$ we have $f(0+)=0$. For all $\epsilon\gt0$, as long as $0 \lt x \lt\delta=\epsilon^{\frac 1 a}$, $\lvert f(x) \rvert\lt\epsilon$. Therefore $f(0+)=0$.

(b) For $f$ to be differentiable at $0$, its left and right derivative must both exist and be the same. It can be shown that $f'(0-)=0$. Therefore we need $f'(0+)=0$.
Since differentiability implies continuity, $a\gt0$. We also have that if $x\gt0$, $f'(x)=ax^{a-1}$. Hence $$f'(0+)=\lim_{x\to0^+}ax^{a-1}$$ We has already shown that the above limit only exists and is zero if $a-1\gt0$. Hence $a\gt1$.

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$$\lim_{0^-}f=0=f(0)$$ So,

$ f $ is continuous at $ 0 \; \iff $ $$\lim_{0^+}f=\lim_{0^+}e^{a\ln(x)}=0$$

$$\iff \;\; a>0$$

$ f$ is differentiable at $ 0 \;\iff$

$$\lim_{0^+}e^{(a-1)\ln(x)}=0$$

$$\iff a>1$$

The derivative is defined by

$$f'(x)=0 \text{ if } x\le 0$$ $$f'(x)=ax^{a-1} \text{ if } x>0$$

It is continuous at $\Bbb R$.

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If $a>0$, the function is continuous. On the other hand $$\lim_{h\to 0} \frac{(0+h)^a-0^{a}}{h}=\lim_{h \to 0} \frac{h^{a} } {h}=\lim_{h \to 0 } h^{a-1}$$ you need that this limit exists, and this limit exists if $a>1$.

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