Excercise 3
Consider the function
\(\begin{equation*} f(x)= \left\lbrace \begin{array}{l} \exp x, if x\leq 0, \\ \cos(x)+x, if x > 0. \end{array} \right. \end{equation*}\)
\((a)\) Show that \(f\) is continuously differentiable at \(x \neq 0\) arbitrarily many times.
\((b)\) Show that \(f\) is (once) continuously differentiable at \(x=0\).
\((c)\) Is \(f\) twice continuously differentiable at \(x=0\)?
Solution
\((a)\) The functiom \(\exp x\) and \(x+\cos x\) are arbitrary often continuously differentiable on the whole of \(\mathbb{R}\) and thus also \(f\) as an arbitrarily often static combination of both.
\((b)\) Further is for \(x\neq 0\)
\(\begin{equation*} f'(x)= \left\lbrace \begin{array}{l} \exp x, if x<0, \\ 1-\sin x, if x > 0. \end{array} \right. \end{equation*}\)
So it applies
\(\lim_{x\rightarrow-0} f'(x) = 1 =\lim_{x\rightarrow+0} f'(x) .\)
The derivation \(f'\) is therefore in \(x=0\) , f is then continuously differentiable in \(x_{0}=0\).
\((c)\) One calculates for \(x\neq0\)
\(\begin{equation*} f''(x)= \left\lbrace \begin{array}{l} e^{x}, if x<0, \\ -\cos x, if x > 0. \end{array} \right. \end{equation*}\)
Here is
\(\lim_{x\rightarrow-0} f'(x) = 1\neq -1 =\lim_{x\rightarrow+0} f''(x).\)
It means that\(f''\) in\( x_{0}=0\) cannot be continuously supplemented and therefore \(f \)has a continuous second derivative!