How to show a function is twice continuously differentiable?

[docnet]

Homework Statement

show $f(x)\in C^2(\mathbb{R}$

Relevant Equations

$$f(x)=\begin{cases} (x+1)^4 & x<-\frac{1}{2} \\ 2x^4-\frac{3}{2}x^2+\frac{5}{16} & -\frac{1}{2}\leq x \end{cases}$$

this seems to come up frequently in undergrad math classes so it is worth asking, what is the simplest and most efficient way to show $f(x)\in C^2(\mathbb{R})$
given $$f(x)=\begin{cases} (x+1)^4 & x<-\frac{1}{2} \\ 2x^4-\frac{3}{2}x^2+\frac{5}{16} & -\frac{1}{2}\leq x \end{cases}$$
And what is the most through and concrete way to show it?
one could compute $f(x)$, $f'(x)$ and $f''(x)$ at $x=-\frac{1}{2}$ to show whether they exist and are continuous. and polynomial functions are $\in C^\infty$ so we only have to consider the boundary. is this sufficient?

 

[PeroK]

That particular function is not continuous at $x = 0$.

 

[docnet]

That particular function is not continuous at x=0.

Apologies sir, there were errors but I fixed them. $f(x)$ should be continuous now.

 

[PeroK]

Apologies sir, there were errors but I fixed them. $f(x)$ should be continuous now.

Apologies sir, there were errors but I fixed them. $f(x)$ should be continuous now.

If you have the following function:
$$f(x)=\begin{cases} g(x) & x \le a \\ h(x) & x > a \end{cases}$$ where $g(x)$ and $h(x)$ are continuously differentiable functions on $\mathbb R$ (or, at least, on an interval containing $a$), then:

First, we have to check that $f$ is continuous at $a$. Then, we can differentiate $g$ and $h$ normally. And, if $h'(a) = g'(a)$, then $f$ is continuously differentiable.