Distinguish a corner from a break

Distance from a marker

The graph shows \(d(x)=|x-3|\), the distance of a walker from a marker at \(x=3\). The orange point sits at the input you choose.

The graph has a sharp corner at 3, but the pen does not lift. The left limit and the right limit are both 0, which is the value \(d(3)\), so \(d\) is continuous at 3.

Move x along the left side. The orange line is the tangent, the line that touches the graph at the point, and its slope is −1. On the right side the slope is 1.

At 3 the graph has two slopes, shown by two orange half-lines, so it has no single tangent line.

Two slopes at one point

The graph now shows any corner at an input \(c\), where two straight pieces meet.

The slope of a graph at a point is the slope of the line that the graph follows there. The two orange half-lines show the slope on each side of \(c\).

Turn the right piece with the slider. Unless the right slope is also \(-1\), the half-lines meet at an angle, so the graph has a corner.

A function is differentiable at \(c\) when its graph has one tangent line at \(c\). A corner has two slopes, so the function is not differentiable there, but the pieces still meet, so it is continuous.

Notation

1. Check continuity first. The function is continuous at \(c\) if the left limit, the right limit and the value are the same number.

   \[\lim_{x\to c^-} f(x)=\lim_{x\to c^+} f(x)=\cy{f(c)}\]

2. Then find the slope on the left of \(c\) and the slope on the right of \(c\).

   \[\cy{m_1}\ \text{(left)},\qquad \cy{m_2}\ \text{(right)}\]

3. If the slopes are different, the graph has a corner at \(c\). If they are equal, the two pieces join into one straight line.

   \[m_1\neq m_2\]

4. If \(f\) is differentiable at \(c\), then \(f\) is continuous at \(c\). At a break, \(f\) is not continuous, so it is not differentiable either.

5. The reverse is false, because at a corner \(f\) is continuous but not differentiable.

When it fails

Press Round the corner. The half-lines fade, and the graph becomes a smooth curve through \(c\).

The smooth curve has one tangent line at \(c\), so the function is now differentiable there.

Example 1: equal slopes, but a break

Let \(t(x)=x+1\) for \(x<2\), and \(t(x)=x+3\) for \(x\geq 2\). Is \(t\) continuous at 2? Is \(t\) differentiable at 2?

1. From the left, \(x+1\) heads to 3. From the right, \(x+3\) heads to 5.

   \[\lim_{x\to 2^-} t(x)=\cy{3},\qquad \lim_{x\to 2^+} t(x)=\cy{5}\]

2. The limits differ, so \(t\) has a jump at 2 and is not continuous there. The green bar shows the rise of 2 across the gap.

3. Both pieces have slope 1, but equal slopes are not enough. Because \(t\) is not continuous at 2, it is not differentiable there.

Example 2: a corner that looks like a break

Let \(s(x)=x-1\) for \(x\leq 3\), and \(s(x)=11-3x\) for \(x>3\). Is \(s\) continuous at 3? Is \(s\) differentiable at 3?

1. From the left, \(x-1\) heads to 2. From the right, \(11-3x\) heads to 2. The value is \(s(3)=2\).

   \[\lim_{x\to 3^-} s(x)=\cy{2},\qquad \lim_{x\to 3^+} s(x)=\cy{2}\]

2. A tempting wrong move: The graph turns sharply at 3, so the pen must lift and \(s\) is not continuous.

   \[\text{sharp turn}\ \Rightarrow\ \text{break}\]

   Both pieces reach the red dot at (3, 2), because the left limit, the right limit and the value are all 2. A pen traces the corner without lifting.

3. The function \(s\) is continuous at 3, but its slope is 1 on the left and \(-3\) on the right. The slopes differ, so \(s\) has a corner at \((3,2)\) and is not differentiable at 3.

Practice

Let \(u(x)=|2x+6|\). Find the input of the corner and the slope on each side, and say whether \(u\) is continuous there.

The point of the V is where the expression inside the absolute value is 0. On each side of it, write u without the bars.

Questions

The graph shows a function F made of straight pieces. It has a sharp corner at one input and a jump at another.

1. At which input is F continuous but not differentiable?

   • \(-1\)
   • 2
   • 0
   • Both \(-1\) and 2
   Answer

   \(-1\)

2. What are the slopes of the graph on the left of \(-1\) and on the right of \(-1\)?

   • \(-1\) and 1
   • 1 and 1
   • \(-1\) and \(-1\)
   • 0 and 1
   Answer

   \(-1\) and 1

3. Which function is continuous at \(x=0\) but not differentiable at \(x=0\)?

   • \(f(x)=2|x|-3\)
   • \(f(x)=2x-3\) for \(x<0\), and \(f(x)=2x+3\) for \(x\geq 0\)
   • \(f(x)=2x-3\)
   • \(f(x)=|x-1|\)
   Answer

   \(f(x)=2|x|-3\)

Summary

At a corner the two pieces meet at one height, so the function is continuous there.

The slopes on the two sides differ, so the graph has no single tangent line and the function is not differentiable.

When a graph has a sharp point or a gap, compare the heights for continuity and the slopes for differentiability.

This is one of five skills in topic 1.10 of Which & Why: AP Calculus.