Use the correct logical implication

Two labels under one graph

The two labels under the graph say whether f is continuous at x = 2 and whether f has a derivative there. Each label is green for yes and red for no.

Slide "Size of the corner" up. The graph is still joined, so the first label stays green, but it has a corner, so the second label turns red.

Slide the corner back to 0, and slide "Size of the jump" up. The graph breaks at 2, and the red segment shows the size of the jump. Both labels turn red.

Try other positions of the sliders. The second label is never green while the first is red, because a graph with a derivative at 2 is continuous at 2.

One set inside another

The small ellipse holds the functions that are differentiable at a, which means that they have a derivative there. The large ellipse holds the functions that are continuous at a.

The small ellipse lies inside the large one, so every function that is differentiable at a is also continuous at a. This is the theorem.

Press "Corner". A corner is continuous but has no derivative, so continuous does not imply differentiable.

Press "Jump". A jump is not continuous, so it lies outside both ellipses and is not differentiable either.

Notation

1. If \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\). This statement is the theorem. In the graph, \(a = 2\).

   \[\begin{aligned} &\text{differentiable at } a \\ &\quad\Longrightarrow\; \text{continuous at } a \end{aligned}\]

2. The converse swaps the two parts. It is false, because a corner is continuous and has no derivative.

   \[\begin{aligned} &\text{continuous at } a \\ &\quad\not\Longrightarrow\; \text{differentiable at } a \end{aligned}\]

3. The contrapositive swaps the two parts and negates both, so it says the same as the theorem and is true. At a break, \(f'(a)\) does not exist.

   \[\begin{aligned} &\cy{\text{not continuous at } a} \\ &\quad\cy{\Longrightarrow\; \text{not differentiable at } a} \end{aligned}\]

4. A counterexample is a function that satisfies the 'if' part of a statement and fails the 'then' part. One counterexample shows that the statement is false.

When it fails

Without the corner, the graph is continuous at 2 and has a derivative there. With the corner, it is still joined, so it is continuous at 2, but it has no derivative there.

The graph with the corner is a counterexample to the converse.

Example 1: a hole filled with the wrong value

Let \(f(x) = \dfrac{x^2-9}{x-3}\) for \(x \ne 3\), and let \(f(3) = 5\). Is \(f\) differentiable at 3?

1. For \(x \ne 3\), \(x^2 - 9 = (x-3)(x+3)\), so you can cancel \(x - 3\). The function equals \(x + 3\), which approaches 6 as \(x\) approaches 3.

   \[\lim_{x\to 3}\dfrac{x^2-9}{x-3} = \lim_{x\to 3}(x+3) = 6\]

2. The limit is 6 but \(f(3) = 5\), so \(f\) is not continuous at 3.

3. The contrapositive gives the answer without any slope. A function that is not continuous at 3 is not differentiable at 3.

   \[\cy{f'(3) \text{ does not exist}}\]

Example 2: an absolute value

Let \(g(x) = |x^2 - 4|\). A student says that \(g\) is continuous at 2, so \(g'(2)\) exists. Is the student correct?

1. The function \(g\) is continuous at 2, because \(x^2 - 4\) approaches 0 and \(g(2) = 0\).

   \[\lim_{x\to 2}|x^2-4| = 0 = g(2)\]

2. A tempting wrong move: Because \(g\) is continuous at 2, it has a derivative at 2.

   \[\cr{\text{continuous}\Longrightarrow\text{differentiable}}\]

   This is the converse of the theorem, and it is false. The graph has a corner at \((2, 0)\), and the red half-lines have slopes −4 and 4.

3. Continuity is needed for a derivative, but it does not give one. To the left of 2, \(g(x) = 4 - x^2\) has slope −4. To the right, \(g(x) = x^2 - 4\) has slope 4.

   \[\cy{g'_-(2) = -4 \ne 4 = g'_+(2)}\]

4. The derivative \(g'(2)\) does not exist, so the student is not correct.

Practice

Say whether each statement is true or false. Give a counterexample for each false statement. (a) If \(f\) is differentiable at 5, then \(f\) is continuous at 5. (b) If \(f\) is continuous at 5, then \(f\) is differentiable at 5. (c) If \(f\) is not continuous at 5, then \(f\) is not differentiable at 5. (d) If \(f\) is not differentiable at 5, then \(f\) is not continuous at 5.

Compare each statement with the theorem. Use the corner slider to test statements (b) and (d).

Questions

Questions 1 and 2 use the two labels from the start of this page. Question 3 needs a calculation.

1. The graph has a corner at x = 2 and no break. Which statement does this graph show to be false?

   • If \(f\) is continuous at \(a\), then \(f\) is differentiable at \(a\).
   • If \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\).
   • If \(f\) is not continuous at \(a\), then \(f\) is not differentiable at \(a\).
   Answer

   If \(f\) is continuous at \(a\), then \(f\) is differentiable at \(a\).

2. The graph now has a jump at \(x = 2\). What follows about \(f'(2)\)?

   • It does not exist, because \(f\) is not continuous at 2.
   • It exists, because each piece of the graph has a slope.
   • It cannot be decided without the slopes of the two pieces.
   Answer

   It does not exist, because \(f\) is not continuous at 2.

3. Let \(h(x) = x^2 + 1\) for \(x \le 1\) and \(h(x) = 4x - 2\) for \(x > 1\). Which statement is correct?

   • \(h\) is continuous at 1 but not differentiable at 1.
   • \(h\) is differentiable at 1, because it is continuous at 1.
   • \(h\) is not continuous at 1, so it is not differentiable at 1.
   • \(h\) is differentiable at 1, because each piece is differentiable.
   Answer

   \(h\) is continuous at 1 but not differentiable at 1.

Summary

You use the theorem in the true direction: a derivative at a point forces continuity there.

The converse fails because a corner is continuous and has no derivative, and one counterexample is enough to show that.

At a break, the contrapositive rules out a derivative, and where the graph is joined, you compare the slopes on each side.

This is one of five skills in topic 2.4 of Which & Why: AP Calculus.