Prove that $f(x) = x_{+}^k$ is continuously differentiable if $k$ is an integer greater than 1. First time doing one of these problems and just want a sanity check.
Define
$$x_{+}=\begin{cases}
x&x \geq 0\\\\
0& x < 0
\end{cases}$$

Prove that $f(x) = x_{+}^k$ is continuously differentiable if $k$ is
an integer greater than 1.

To prove that $f$ is continuously differentiable, we need to show that for every point, a derivative exists and that such a derivative is continuous. Consider an $x_0 < 0$. Then, we know that $$\frac{f(x)-f(x_0)}{x-x_0} = \frac{0-0}{x-x_0} = 0$$ and so $f'(x_0) = 0$ because $$\left\lvert \frac{f(x)-f(x_0)}{x-x_0} - f'(x_0)\right\rvert \leq 1/m \Longleftrightarrow \left\lvert 0 - f'(x_0)\right\rvert \leq 1/m \Longleftrightarrow f'(x_0) = 0$$ Now, consider an $x_0 > 0$. Then, we know that
\begin{align*}
    f'(a) &= \lim_{x \to a}\frac{x^n-a^n}{x-a} \\
    &= \lim_{x \to a}\frac{(x-a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-3}x^2 + a^{n-2}x + a^{n-1})}{x-a} \\
    &= \lim_{x \to a}(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-3}x^2 + a^{n-2}x + a^{n-1}) \\
    &= a^{n-1} + aa^{n-2} + a^2a^{n-3} + \cdots + a^{n-3}a^2 + a^{n-2}a + a^{n-1} \\
    &= na^{n-1}
\end{align*}
Now, consider $x_0 = 0$. The derivative from the left is trivially 0, and the derivative from the right is $\lim_{x \to 0^{+}}kx^{k-1} = k(\lim_{x \to 0} x^{k-1}) = 0$. Hence, every point has a derivative. (By the way, we need the assumption that $k > 1$ because then the derivative from the right would be 1!)
To prove that the derivative is continuous, we know that the zero function for $x < 0$ is trivially continuous. We also know that for $x > 0, nx^{n-1}$ is continuous because continuity is maintained by scalar multiplication as well as products of functions. For $x = 0$, we know that $f(x) = 0$, and because the left side is always 0, we only need to worry about continuity from the right. Well, choose an arbitrary $1/m$. We know that $f(x) = kx^{k-1}$. Consider the interval $(0,1/km)$. We know that the largest number in this interval is $k(\frac{1}{km})^{k-1} = k(\frac{1}{k^{k-1}m^{k-1}}) = \frac{1}{k^{k-2}m^{k-1}} < 1/m$. Hence, the derivative is continuous.
 A: The OP's solution is valid and they can now try there hand working on the following more general theory:
Lemma 1: If $f$ is continuous on an interval $I_1$ and if $I_2 \subset I_1$ is a subinterval,
then the restriction of $f$ to $I_2$ is also continuous.
See also Properties of restrictions.
Lemma 2: If $f: [a,b] \to \Bbb R$ is continuous and and $g: [b,c] \to \Bbb R$ is continuous and $f(b) = g(b)$, then the function $h: [a,c] \to \Bbb R$ defined by 'pasting' $f$ and $g$ together is also continuous.
See also Piecewise functions and Pasting lemma.
A: I think you already did all the leg-work, but if I may summarize:

*

*We need to prove that for $k>1$:
$$f'(x)=\begin{cases}
kx^{k-1} &\text{ if} &x\geq 0\\
0 &\text{ if} &x< 0
\end{cases}
$$


*$f'(x)$ is continuous in al it's domain (which in only non-trivial at $x_0=0$).
For 1, we can use OP's work.
If we want to go for and epsilon-delta proof for 2, we can take $\delta(\epsilon)=\left(\frac{\epsilon}{k}\right)^{\frac{1}{k-1}}$ and notice that for all $\epsilon$ and for $x_0=0$ we have that:
$$|x-x_0|=|x|<\delta(\epsilon)=\left(\frac{\epsilon}{k}\right)^{\frac{1}{k-1}}$$
