My professor emphasized that:
- Differentiability implies continuity and
- Continuity is required for differentiability.
Since a function like $\frac 1 x$ is differentiable but not continuous, I thought my professor simply forgot to say that the 2 rules only apply at a point, not an interval.
However, in the textbook, we were given the following questions and the corresponding solutions:
- If $f$ is differentiable and $f(-1)=f(1),$ then there is a number $c$ such that $|c|<1$ and $f'(c)=0.$ (true)
My solution: consider $f=\frac 1 {x^2}$, therefore it is false.
- If $f'(x)$ exists and is nonzero for all $x,$ then $f(1)\neq f(0).$ (true)
My solution: consider $f=\frac 1 {(x-0.5)^2}$, therefore it is false.
The textbook's answer only makes sense if differentiability implies continuity on an interval. So does differentiability imply continuity on an interval or is the textbook wrong?
Best Answer
The functions $f(x) = 1/x$ and $f(x) = 1/x^2$ are not defined in $0$. So in particular it makes no sense to think about continuity or differentiability at $0$. Both your statement hold only on intervals.
Differentiability does not imply continuity on an interval! Consider the somewhat artificial functions defined as $0$ on the rationals and $x^2$ on the irrationals. It is continuous and differentiable at $0$ and neither continuous nor differentiable on $\mathbb{R} \setminus \{0\}$.
Edit: I think I misunderstood the "on an interval" part. Anyway, the implication is pointwise.