For f to be continuous at some point x = c, you require that
• f(c) exists and is finite, and
• the limits of f as x approaches c from either side match, and their value must be f(c)
For f to be differentiable at x = c, you require that
• f is continuous at c (i.e. the above conditions are all met), and
• the derivative f ' is continuous at c
Without having the graph of f ', you can assess whether the last condition is met by some function by mentally tracing a tangent line to the graph as you get closer to c. If the slope of the tangent doesn't change, then the function is differentiable. This is usually accompanied by jumps or sharp corners in the graph.
Some examples:
• every polynomial is both continuous and differentiable
• the absolute value function |x| is continuous but not differentiable at x = 0
• 1/x is neither continuous nor differentiable at x = 0
(a) Neither continuous nor differentiable
Why? f has a vertical asymptote at x = -2 and f (-2) does not exist. f is not continuous, and therefore not differentiable.
(b) Neither continuous nor differentiable
Why? From the left, f is approaching 2.5, while from the right, it's approaching 5.
(c) Neither continuous nor differentiable
Why? The limits from either side of x = 2 match and are equal to 0.5, but f (2) itself does not exist (which is indicated by the point being a hollow circle).
If the circle was instead filled in, so that f (2) = 0.5, then f would be continuous there, but still not differentiable. Notice the sharp corner. Or, using the tangent-line analysis, the slope of the tangent to the left of x = 2 is negative, but to the right it would be positive.
(d) Both continuous and differentiable
Why? f doesn't have any special features at x = 3 that would suggest it's not continuous nor differentiable.
(e) Neither continuous nor differentiable
Why? Similar reasoning as in (a). There's another vertical asymptote, but the graph shows f (4) = 1. However, the limits from either side of x = 4 are positive infinity, not 1.
