Unit 2 · Topic 2.4 Beta

Differentiability and Continuity

A function is differentiable at a when f′(a) exists.

Practice 3: Justification

Part 1 · Hook

Why this matters

A ramp that meets the floor at a sharp angle jolts a skateboard; a ramp that curves smoothly into the floor does not. Both are connected, with no gap to jump. Calculus draws the same line: a graph can be connected, continuous, and still have a point where no single slope exists.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A function $f$ is continuous at $x = a$ when

$\lim_{x \to a} f(x)$ $= f(a)$
$f(a)$ is defined
$\lim_{x \to a} f(x)$ exists
$f(a) = 0$

Show the answer

All three conditions at once: $f(a)$ is defined, the limit exists, and they are equal.

Correct: $\lim_{x \to a} f(x)$ $= f(a)$ :
$f(a)$ is defined:
$\lim_{x \to a} f(x)$ exists:
$f(a) = 0$ :

2. For $f(x) = x^2$ , the definition of the derivative gives

$f'(x) = 2x$
$f'(x) = x$
$f'(x) = 2x + h$
$f'(x) = x^2$

Show the answer

The difference quotient simplifies to $2x + h$ , whose limit as $h \to 0$ is $2x$ .

Correct: $f'(x) = 2x$ :
$f'(x) = x$ :
$f'(x) = 2x + h$ :
$f'(x) = x^2$ :

3. If $\lim_{x \to 1^-} f(x) = 3$ and $\lim_{x \to 1^+} f(x) = 5$ , then $\lim_{x \to 1} f(x)$

does not exist
is 4
is 3
is 5

Show the answer

A two-sided limit exists only when the one-sided limits are equal.

Correct: does not exist:
is 4:
is 3:
is 5:

Part 4 · See it

See it first

Four small graphs. A corner: y = |x| + 0.5, a V shape with slope minus 1 on the left of 0 and 1 on the right. A cusp: y = x to the two-thirds, a sharp point at the origin where the slopes become infinitely steep in opposite directions. A vertical tangent: y = cube root of (x minus 1), with a vertical tangent line x = 1 drawn dashed. A jump: a line ending with a closed dot at (1, 1.5) and restarting with an open dot at (1, 0). The first three are continuous but not differentiable at the marked point; the jump is not continuous, so not differentiable. — Three continuous graphs that are not differentiable at one point (a corner, a cusp, a vertical tangent) and a jump, which is not continuous and so not differentiable. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f'(a)$ exists $f(a + h) - f(a)$ $= \frac{f(a + h) - f(a)}{h} \cdot h$ tends to $f'(a) \cdot 0 = 0$
$f(a + h) - f(a) \to 0$ as $h \to 0$ $\lim_{x \to a} f(x)$ $= f(a)$ : $f$ is continuous at $a$
The left and right difference quotients approach different numbers (a corner)the two-sided limit does not exist and $f'(a)$ does not exist, even though $f$ may be continuous
The difference quotients grow without bound (a cusp or a vertical tangent)there is no finite slope, so $f'(a)$ does not exist

Part 6 · Key ideas

Key ideas

$f$ is differentiable at $a$ when $f'(a)$ exists: the limit of the difference quotient is a finite number.
Differentiable implies continuous. In a justification, say it: " $f$ is differentiable at $a$ , so $f$ is continuous at $a$ ."
Continuous does NOT imply differentiable. A continuous graph fails to have a derivative at a corner (one-sided slopes differ), a cusp (slopes become infinite in opposite directions) or a vertical tangent line.
Not continuous implies not differentiable: a jump, hole or asymptote rules out a derivative.
For a piecewise function at the break: check continuity first, then compare the left and right slopes.

Part 7 · Misconception

A common mistake

The wrong idea: The graph of $|x - 3|$ has no break at $x = 3$ , so it has a derivative there.

What actually happens: No break means continuous, which is necessary but not enough. At $x = 3$ the slope is $-1$ from the left and 1 from the right, so the difference quotient has no limit: $|x - 3|$ is continuous but not differentiable at 3.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of a function $g$ on $[-3, 4]$

The graph of

g

is made of a line segment on

[-3, 0]

, part of the parabola

y = \frac{x^2}{2}

[0, 2]

, and part of the curve

y = 1 + \sqrt[3]{x - 3}

(2, 4]

1. At which values of $x$ in the open interval $(-3, 4)$ is $g$ not differentiable? Select every one.

$x = -2$
$x = 0$
$x = 1$
$x = 2$
$x = 3$
$x = 3.5$

Show the answer

A corner at $x = 0$ , a jump at $x = 2$ and a vertical tangent at $x = 3$ . Everywhere else the graph is smooth.

$x = -2$ : The graph is a straight line here, with slope $-1$ .
Correct: $x = 0$ : A corner: slope $-1$ on the left, 0 on the right.
$x = 1$ : The parabola is smooth here, with slope 1.
Correct: $x = 2$ : A jump: $g$ is not continuous at 2, so it is not differentiable there.
Correct: $x = 3$ : The curve is vertical at $(3, 1)$ : the difference quotients grow without bound.
$x = 3.5$ : The curve is smooth here, with a finite slope.

2. Which is true of $g$ at $x = 2$ ?

It is neither continuous nor differentiable.
It is continuous but not differentiable.
It is differentiable at 2 but not continuous at 2.
It is both continuous and differentiable.

Show the answer

$\lim_{x \to 2^-} g(x) = 2$ and $\lim_{x \to 2^+} g(x) = 0$ , so $g$ is not continuous at 2, and a function that is not continuous cannot be differentiable.

Correct: It is neither continuous nor differentiable.: A jump breaks continuity, and with it differentiability.
It is continuous but not differentiable.: The one-sided limits are 2 and 0, so $g$ is not continuous at 2.
It is differentiable at 2 but not continuous at 2.: Differentiable implies continuous, so this combination is impossible.
It is both continuous and differentiable.: There is a jump at $x = 2$ .

3. Why is $g$ not differentiable at $x = 3$ ?

The tangent line at $(3, 1)$ is vertical: the slopes grow without bound.
$g(3)$ is not defined, so there is no point to take a slope at.
$g$ is not continuous at $x = 3$ .
The graph has a corner at $x = 3$ , with two different finite slopes.

Show the answer

Near $x = 3$ the curve $1 + \sqrt[3]{x - 3}$ is steeper and steeper; the slopes have no finite limit.

Correct: The tangent line at $(3, 1)$ is vertical: the slopes grow without bound.: A vertical tangent line has no finite slope.
$g(3)$ is not defined, so there is no point to take a slope at.: $g(3) = 1$ ; the point is on the graph.
$g$ is not continuous at $x = 3$ .: The curve passes through $(3, 1)$ without a break, so $g$ is continuous there.
The graph has a corner at $x = 3$ , with two different finite slopes.: Both sides become infinitely steep in the same direction; there are no two finite slopes.

4. Which statement best explains why $g$ is not differentiable at $x = 0$ ?

$g$ is continuous at 0, but the difference quotient approaches $-1$ from the left and 0 from the right.
$g$ is not continuous at 0, because two different formulas meet there.
$g(0) = 0$ , and a function has no derivative where its value is 0.
The graph is level at $x = 0$ , so $g'(0)$ does not exist.

Show the answer

On the left the graph is the line $y = -x$ (slope $-1$ ); on the right the parabola $\frac{x^2}{2}$ has slope $x$ , which is 0 at $x = 0$ . Two different one-sided slopes make a corner.

Correct: $g$ is continuous at 0, but the difference quotient approaches $-1$ from the left and 0 from the right.: Continuous, but the one-sided slopes differ: a corner.
$g$ is not continuous at 0, because two different formulas meet there.: Both formulas give 0 at $x = 0$ , so $g$ is continuous; a change of formula alone breaks nothing.
$g(0) = 0$ , and a function has no derivative where its value is 0.: A function can be 0 and still have a derivative; $x^2$ does at $x = 0$ .
The graph is level at $x = 0$ , so $g'(0)$ does not exist.: A level tangent line would mean $g'(0) = 0$ , which exists; here no single tangent line exists.

5. Which statement is true for every function $f$ ?

If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ .
If $f$ is continuous at $a$ , then $f$ is differentiable at $a$ .
If $f$ is not differentiable at $a$ , then $f$ is not continuous at $a$ .
If $f(a)$ is defined, then $f$ is differentiable at $a$ .

Show the answer

Differentiability implies continuity.

Correct: If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ .: The theorem: a derivative at $a$ forces continuity at $a$ .
If $f$ is continuous at $a$ , then $f$ is differentiable at $a$ .: The converse is false: $|x|$ is continuous at 0 but not differentiable there.
If $f$ is not differentiable at $a$ , then $f$ is not continuous at $a$ .: False: $|x|$ is not differentiable at 0 yet is continuous there.
If $f(a)$ is defined, then $f$ is differentiable at $a$ .: A value at $a$ does not even guarantee continuity, let alone a derivative.

6. A function $f$ is continuous at $x = a$ . What can you conclude about $f'(a)$ ?

$f'(a)$ may or may not exist.
$f'(a)$ exists, because continuity guarantees a slope.
$f'(a)$ does not exist.
$f'(a) = f(a)$ .

Show the answer

Continuity is necessary for a derivative but not sufficient: $x^2$ (differentiable at 0) and $|x|$ (not differentiable at 0) are both continuous at 0.

Correct: $f'(a)$ may or may not exist.: Continuity alone does not decide it.
$f'(a)$ exists, because continuity guarantees a slope.: A corner, cusp or vertical tangent can be continuous with no derivative.
$f'(a)$ does not exist.: Many continuous functions, such as $x^2$ , do have derivatives.
$f'(a) = f(a)$ .: A slope and a height are different quantities.

7. A student writes: " $f$ is not differentiable at $x = 2$ , so $f$ is not continuous at $x = 2$ ." Which response is correct?

The reasoning is wrong: $f(x) = |x - 2|$ is continuous at 2 but not differentiable there.
The reasoning is right: differentiability and continuity mean the same thing.
The reasoning is right: it is the theorem that differentiable functions are continuous.
The reasoning is wrong: a function that is not differentiable is continuous.

Show the answer

The theorem runs one way: differentiable implies continuous. Its converse, and the student's statement, can fail.

Correct: The reasoning is wrong: $f(x) = |x - 2|$ is continuous at 2 but not differentiable there.: One counterexample shows the claim is false.
The reasoning is right: differentiability and continuity mean the same thing.: They differ: corners, cusps and vertical tangents are continuous points with no derivative.
The reasoning is right: it is the theorem that differentiable functions are continuous.: The student reversed the theorem; "not differentiable" says nothing about continuity.
The reasoning is wrong: a function that is not differentiable is continuous.: A jump is not differentiable and not continuous either; no conclusion either way.

Part 9 · Summary

Summary

A function is differentiable at

a

when

f'(a)

exists. Differentiability implies continuity, but not the other way round: corners, cusps and vertical tangents are continuous points with no derivative, and any discontinuity also rules a derivative out. At a break in a piecewise function, check continuity, then compare the one-sided slopes.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of a function gg on [−3,4][-3, 4]

Summary

What comes next

Connections

The graph of a function $g$ on $[-3, 4]$