How to show a function is twice continuously differentiable?

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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?
 
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  • #2
PeroK
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That particular function is not continuous at ##x = 0##.
 
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That particular function is not continuous at x=0.

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

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