Differentiability and Continuity

The proof is short. For

x \ne a

f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a).

x \to a

, the fraction approaches

f'(a)

and

x - a

approaches

0

, so

f(x) - f(a)

approaches

f'(a) \cdot 0 = 0

. That says

\lim_{x \to a} f(x) = f(a)

, which is continuity at

a

Equivalently, a function that isn't continuous at

a

isn't differentiable there. In particular, if

a

isn't in the domain of

f

, it isn't in the domain of

f'

. The converse is false: a continuous function can fail to have a derivative.

A jump

A corner

A cusp

A vertical tangent (dashed)

Corners

Example 1

Let

f(x) = |x^2 - 4|

. Show that

f

is continuous at

x = 2

but not differentiable there.

The function is continuous everywhere, as the absolute value of a polynomial, and

f(2) = 0

. When

x \ge 2

, the formula is

f(x) = x^2 - 4

. When

x

is just below

2

x^2 - 4

is negative, so

f(x) = 4 - x^2

. For

h > 0

\begin{aligned} &\frac{f(2 + h) - f(2)}{h} \\ &\qquad = \frac{4h + h^2}{h} = 4 + h \to 4, \end{aligned}

and for

-4 < h < 0

\begin{aligned} &\frac{f(2 + h) - f(2)}{h} \\ &\qquad = \frac{-4h - h^2}{h} = -4 - h \to -4. \end{aligned}

The one-sided derivatives are

4

and

-4

, so

f'(2)

doesn't exist. The graph has a corner at

(2, 0)

Zooming in shows the difference. Near

x = 1

, where

f(x) = 4 - x^2

is differentiable, the graph looks more and more like a line. Near

x = 2

it stays a corner at every magnification.

y = ∣x² − 4∣ on [−3, 3]

Zoomed in at (1, 3): nearly a line

Zoomed in at (2, 0): still a corner

Cusps and vertical tangents

Example 2

(a) Let

g(x) = x^{2/3}

. Show that

g'(0)

doesn't exist, and describe the graph at the origin.

(b) Let

k(x) = \sqrt[3]{x - 1}

. Show that

k'(1)

doesn't exist, and describe the graph at

(1, 0)

(a)

g

is continuous, and

g(0) = 0

. The difference quotient is

\frac{g(h) - g(0)}{h} = \frac{h^{2/3}}{h} = \frac{1}{h^{1/3}}.

h \to 0^+

it approaches

\infty

, and as

h \to 0^-

the cube root of

h

is negative, so it approaches

-\infty

. The graph has a cusp at the origin.

(b)

k

is continuous, and

k(1) = 0

. The difference quotient is

\frac{k(1 + h) - k(1)}{h} = \frac{h^{1/3}}{h} = \frac{1}{h^{2/3}}.

Since

h^{2/3} = \left(h^{1/3}\right)^2

is positive for

h \ne 0

, the quotient approaches

\infty

from both sides. The graph has a vertical tangent, the line

x = 1

A cusp at the origin

A vertical tangent at (1, 0)

Where two formulas meet

At a point where a piecewise function changes formulas, check continuity first. If

f

isn't continuous there, it isn't differentiable. If it is, compare the one-sided derivatives.

Example 3

Decide whether each function is differentiable at

x = 1

\begin{aligned} p(x) &= \begin{cases} x^2, & x \le 1 \\ 2x - 1, & x > 1 \end{cases} \\[4pt] q(x) &= \begin{cases} x^2, & x \le 1 \\ x, & x > 1 \end{cases} \end{aligned}

Both are continuous at

1

, since each formula gives

1

there. The left-hand derivatives agree, because both functions are

x^2

for

x \le 1

\begin{aligned} \lim_{h \to 0^-} \frac{(1 + h)^2 - 1}{h} &= \lim_{h \to 0^-} (2 + h) \\ &= 2. \end{aligned}

On the right,

\frac{2(1 + h) - 1 - 1}{h} = 2

for

p

and

\frac{(1 + h) - 1}{h} = 1

for

q

. So

p'(1) = 2

, while

q

has a corner: its one-sided derivatives are

2

and

1

The quotient

\frac{(1 + h)^2 - 1}{h} = 2 + h

holds for

h > 0

too, so

x^2

has derivative

2

1

. Its tangent line there is

y = 1 + 2(x - 1) = 2x - 1

. So

p

carries on along the tangent line, and the join is smooth.

p: the parabola continues along its own tangent line.

q: the line y = x is less steep, so the graph bends at (1, 1).

Example 4

Find the constants

a

and

b

that make

f

differentiable at

x = 2

f(x) = \begin{cases} ax^2 + b, & x \le 2 \\[2pt] \dfrac{4}{x}, & x > 2 \end{cases}

Continuity at $2$ . The left formula gives

f(2) = 4a + b

, and the right formula approaches

\frac{4}{2} = 2

. So

4a + b = 2

Equal one-sided derivatives. With

f(2) = 2

, the right-hand quotient is

\begin{aligned} &\frac{1}{h}\left(\frac{4}{2 + h} - 2\right) \\ &\qquad = \frac{4 - 4 - 2h}{h(2 + h)} = \frac{-2}{2 + h} \to -1. \end{aligned}

The left-hand quotient is

\begin{aligned} &\frac{a(2 + h)^2 + b - (4a + b)}{h} \\ &\qquad = \frac{4ah + ah^2}{h} \\ &\qquad = 4a + ah \to 4a. \end{aligned}

4a = -1

, which gives

a = -\frac{1}{4}

and then

b = 2 - 4a = 3

. The function is

3 - \frac{x^2}{4}

up to

x = 2

and

\frac{4}{x}

after it.

With a = −¼ and b = 3, the two pieces meet at (2, 2) with the same slope, −1.

A function that oscillates

Example 5

Let

f(x) = x\sin\frac{1}{x}

and

g(x) = x^2\sin\frac{1}{x}

for

x \ne 0

, with

f(0) = g(0) = 0

. Both are continuous at

0

, by the squeeze theorem. Decide whether

f'(0)

and

g'(0)

exist.

For

f

, the difference quotient at

0

\frac{f(h) - f(0)}{h} = \frac{h\sin(1/h)}{h} = \sin\frac{1}{h},

which keeps swinging between

-1

and

1

h \to 0

and has no limit. So

f'(0)

doesn't exist. This failure is none of the four above: the secant lines from the origin keep turning, between slopes

-1

and

1

, and never settle.

For

g

, the quotient is

\frac{h^2\sin(1/h)}{h} = h\sin\frac{1}{h}

, which lies between

-|h|

and

|h|

. By the squeeze theorem it approaches

0

, so

g'(0) = 0

y = x sin(1/x) between y = ±x

y = x² sin(1/x) between y = ±x²: its tangent line at the origin is the x-axis.