I am learning Khan Academy Calculus, specifically how to tell when the graph of a function is differentiable or not. Khan Academy tells me that a point on the graph of is not differentiable when:
But I don't get why 2. and 3. are true. For 2, say we have a vertical tangent line for function 1/x^2
Would it not be continuous here? And therefore, wouldn't there be at least the chance that it's differentiable? Can someone please explain to me why a function is never differentiable at the vertical asymptote?
For 3, why would a function not be differentiable when it has a sharp turn? I don't get this either. Can someone please explain?
$\endgroup$ 22 Answers
$\begingroup$For your given function $f(x)=\frac{1}{x^2}$ this is not differentiable at 0 since the function isn't even continuous there since at 0 it is undefined. It approaches infinity from either side leading to a infinite gradient on either side but AT 0 it does not exist. For (3) I would think a better way of saying this is a 'kink' for example take $f(x)=|x|$. This is not differentiable at 0 since from 0+ we have gradient 1, and -1 from 0-. Or you can think about drawing lots of tangents at the kink, they all look like tangents but are a all different lines so it is not differentiable.
$\endgroup$ 4 $\begingroup$At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$.” isn’t even a statement at all, so it can be neither wrong nor right.
$\endgroup$
