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?
$\endgroup$ 33 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$The other answers are perfectly fine... but here's a proof for why continuity holds:
Suppose that $f$ is differentiable at $x$, then $f'(x)$ exists, so for any $h>0$:
$$f(x+h)-f(x) = \frac{f(x+h)-f(x)}{h} h \to f'(x)\cdot 0$$
as $h\to 0$ because $f'$ is differentiable at $x$.
Therefore, $f(x+h)\to f(x)$ as $h\to 0$ so $f$ is continuous at $x$.
Likewise, if the function is differentiable on an interval then it is continuous on that interval.
$\endgroup$ $\begingroup$A basic theorem states that if a function is differentiable at a point, then it is continuous at that point. The function $f(x)=1/x$ is differentiable and continuous on its domain of definition $\mathbb R\setminus \{0\}$.
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