Continuity

A function can be continuous but not differentiable

A function is continuous at a point if there's no break or jump there, but it's only differentiable at that point if the slope approaching from the left matches the slope approaching from the right. The left-hand derivative is:

Worked example

Find the left-hand derivative of $f(x) = |x|$ at $x = 0$.

Solution: Just to the left of 0, $x < 0$, so $f(x) = |x| = -x$. Differentiating this piece:

So the left-hand derivative at $x=0$ is -1. (The right-hand derivative there is $+1$, since $f(x)=x$ for $x>0$, the two don't match, which is exactly why $|x|$ is continuous at 0 but has a sharp corner there instead of a smooth tangent.)

A function can only be continuous where it's defined

Before asking whether a function is continuous at a point, it has to actually be defined there, so finding a function's domain is usually the first step.

Find the domain of $f(x) = \sqrt{\cos x}$ over one full period, $[0, 2\pi]$.

Solution: A square root is only defined for a non-negative input, so $f(x)$ requires $\cos x \ge 0$. Over $[0, 2\pi]$, cosine is non-negative on the first and last quarters of the circle: