Differentiability and continuity

Warm-up

Three quick questions from earlier lessons.

Discover

$f(x) = |x|$ . Just left of 0 the slope is $-1$ ; just right of 0 it is $1$ . Does the graph have one tangent line at the origin?

No
Yes, since the graph has no break

Key idea

Differentiable means the limit defining $f'(a)$ exists.

If

f

is differentiable at

a

, then

f

is continuous at

a

. The converse is false: a continuous function fails to be differentiable at a corner, a cusp or a vertical tangent. A function is also not differentiable where it is not continuous.

Worked example

Is $f(x) = |x - 2| + 1$ differentiable at $x = 2$ ?

f

is continuous at

2

No jump or hole at 2.

Slope

-1

on the left,

1

on the right

The two pieces are 1 − (x − 2) and 1 + (x − 2).

Not differentiable at

x = 2

: a corner

The one-sided slopes differ.

Your turn, guided

Your turn: $f(x) = x^2$ for $x \le 1$ and $f(x) = 2x - 1$ for $x > 1$ .

The pieces meet at

(1, 1)

Both pieces give 1 at x = 1.

Slope from the left:

2

, from the right:

2

, so

f

is differentiable at

1

The slope of x² at 1 is 2 · 1.

Practice ★

Is $f(x) = |x - -1|$ differentiable at $x = -1$ ?
1. No: the graph has a corner there (slopes −1 and 1).
2. Yes, because $f$ is continuous at $x = -1$ .
3. No, because $f$ is not continuous at $x = -1$ .
Is $f(x) = |x - 1|$ differentiable at $x = 1$ ?
1. No: the graph has a corner there (slopes −1 and 1).
2. Yes, because $f$ is continuous at $x = 1$ .
3. No, because $f$ is not continuous at $x = 1$ .

Watch out!

Practice ★★

Which functions are differentiable at $x = 0$ ? Tick all that are.
1. $x^2 + 8x$
2. $\sin(8x)$
3. $8|x|$
4. $\sqrt[3]{x}$
5. $f(x) = 0$ for $x < 0$ and $f(x) = 8$ for $x \ge 0$

$f(x) = x^2$ for $x \le 4$ and $f(x) = mx - 16$ for $x > 4$ . Which value of $m$ makes $f$ differentiable at $x = 4$ ?
$f(x) = x^2$ for $x \le 5$ and $f(x) = mx - 25$ for $x > 5$ . Which value of $m$ makes $f$ differentiable at $x = 5$ ?

Exit ticket

Three questions to check you have got it.

$f(x) = x^2$ for $x \le 4$ and $f(x) = mx - 16$ for $x > 4$ . Which value of $m$ makes $f$ differentiable at $x = 4$ ?
$f(x) = x^2$ for $x \le 5$ and $f(x) = mx - 25$ for $x > 5$ . Which value of $m$ makes $f$ differentiable at $x = 5$ ?
$f(x) = x^2$ for $x \le 6$ and $f(x) = mx - 36$ for $x > 6$ . Which value of $m$ makes $f$ differentiable at $x = 6$ ?

Summary card

Differentiability

Differentiable ⇒ continuous; not the other way round.
No derivative at a corner, a cusp, a vertical tangent, or a break.
To join two pieces smoothly, match the heights and the slopes.

Watch out: an unbroken graph can still have no tangent line.

About this lesson