Skip to content

Before the midterm

Assuming a continuous function is differentiable

The habit: A derivative at a point forces the graph to be unbroken there, but not the reverse: |x| is unbroken at 0 and has a corner, so no slope.

5 practice questions are built to catch it, in 3 topics.

See it fail in a figure: Where there is no slope, in the definition of the derivative lesson

Practise against it: 4 questions

What it looks like

A question from the definition of the derivative, with the answer this slip gives beside the right one.

Introductory2 marks · 3 min

The function ff is piecewise linear: its graph is made of straight segments joining (−2,1)(-2, 1), (1,7)(1, 7), (3,7)(3, 7) and (5,−1)(5, -1), in that order, as shown. Which statement about its derivative is correct?

A graph made of three straight segments. It climbs from (−2, 1) to (1, 7), runs level from (1, 7) to (3, 7), then falls steeply from (3, 7) to (5, −1). The four joining points are marked.-3-2-1123456-2-112345678
With the slip
f′(0)=2f'(0) = 2, f′(2)=0f'(2) = 0, f′(4)=−4f'(4) = -4, and f′(1)=1f'(1) = 1
Worked through
f′(0)=2f'(0) = 2, f′(2)=0f'(2) = 0, f′(4)=−4f'(4) = -4, and f′(1)f'(1) does not exist

The graph is unbroken at x = 1, but it turns a corner there: slope 2 just to the left, 0 just to the right. No single tangent fits a corner, and an average of the two slopes is not one.

See the worked solution

Where it comes up

Practise against it

Each of these has an answer this slip gives. Work them on paper: if you fall into it, the verdict names it and the question comes back in review.

Start the set