see the attached problem below
My thoughts on (6) are as follows, if we can show that $h$ is continuous on the interval $[0,1]$ then $h$ must be integrable on $[0,1]$ also? my problem is how do i show that a function like this is continuous as it is piecewise?
Problem (7) will just be using the definition of being differentiable i assume? and for problem(8) we will be able to apply the same method as that of in 6 to the derivative of h?
can anyone let me know if i'm along the right lines with my approach
Let $f(x)$ be defined by
$$f(x)=\begin{cases}x^2\cos(1/x^2)&,x\ne 0\\\\0&,x=0\end{cases}$$
For $x\ne 0$, $f(x)$ is the composition of continuous functions and is, therefore, a continuous function.
We also have $\lim_{x\to 0}f(x)=0=f(0)$ and so, $f(x)$ is everywhere continuous and hence Riemann integrable on every closed interval.
To show that $f(x)$ is differentiable everywhere, we first note that $f(x)$ is differentiable for $x\ne0$ since it is composed of differentiable functions. For $x=0$ we find that
$$f'(0)=\lim_{h\to 0}\frac{h^2\cos(1/h^2)-0}{h}=0$$
So, $f'(x)$ exists for all $x$ and can be written
$$f'(x)=\begin{cases}2x\cos(1/x^2)+\frac{2\sin(1/x^2)}{x}&,x\ne 0\\\\0&,x=0\end{cases} \tag 2$$
However, since $f'(x)$ as given in $(2)$ is unbounded in every neighborhood of $x=0$, then it cannot be Riemann integrable there.