Where a Function Is Differentiable

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differentiable at a ⇒ continuous at a

a corner can be continuous and still not differentiable

What to Remember

Differentiable means continuous

If \(f'(a)\) exists, then f is continuous at a. The converse is false.

A corner or a break blocks the derivative

A jump or a hole means f is not continuous, so it is not differentiable. A corner is continuous, but the left-hand slope and the right-hand slope disagree.

A vertical tangent is not a real derivative

If the difference quotient grows without a bound, the derivative is not a real number.

A piecewise function needs two matches

At the boundary, the two formulas must give the same value, and they must give the same derivative. Matching the value alone leaves a corner.

Level 1

Easy

Read continuity and slope from the graph.

Example

The graph is made of straight pieces, with a corner at \(x = 2\). Is f continuous at 2? Is f differentiable at 2?

Graph of f

Both pieces meet at \((2, 0)\), and the point is filled.

f is continuous at 2.

The left slope is \(-1\) and the right slope is 1.

The slopes disagree, so f is not differentiable at 2.

Answer: continuous, not differentiable

Use the graph. Is f continuous at \(x = -2\)? Is it differentiable there?

Graph of f

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The graph approaches height 4 from both sides.

The closed dot is at height 1.

\(\lim_{x \to -2} f(x) = 4 \neq 1 = f(-2)\)

f is not continuous at −2, so it is not differentiable there.

Answer: neither

Use the graph. Find \(f'(0)\), or say it does not exist.

Graph of f

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x = 0 is on the open segment between the hole and the corner.

The slope of the segment from \((-2, 4)\) to \((2, 0)\) is

\(\dfrac{0 - 4}{4} = -1\)

Answer: \(-1\)

If \(f'(4)\) exists, must f be continuous at 4?

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Differentiability at 4 implies continuity at 4.

Answer: yes

If f is continuous at 4, must \(f'(4)\) exist?

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A corner can be continuous and still fail to be differentiable.

Continuity does not imply differentiability.

Answer: no

For \(f(x) = |x - 1|\), is f continuous at \(x = 1\)? Is it differentiable there?

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\(f(1) = 0\)

Both sides approach 0, so f is continuous at 1.

For \(x \lt 1\), \(f(x) = 1 - x\), so the left-hand derivative is \(-1\).

For \(x \gt 1\), \(f(x) = x - 1\), so the right-hand derivative is 1.

The one-sided derivatives are not equal.

Answer: continuous, not differentiable

Level 2

Medium

Say which statement must be true.

Example

If \(f'(2)\) exists, which must be true?

A. f is continuous at \(x = 2\)

B. \(f'(2) = 0\)

C. f is continuous at every real number

D. \(f'\) is continuous at \(x = 2\)

Differentiability at 2 implies continuity at 2.

The derivative need not be 0, and it need not be continuous.

Other x-values are not involved.

Answer: A. f is continuous at x = 2

f is continuous at \(x = 3\). Which must be true?

A. \(f'(3)\) exists

B. \(\lim_{x \to 3} f(x) = f(3)\)

C. \(f'(3) = 0\)

D. \(\lim_{x \to 3} f'(x)\) exists

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Continuity at 3 means the limit equals the function value.

It does not create a derivative.

Answer: B. the limit equals f(3)

Use the graph. At which labeled x-value is f continuous but not differentiable: \(-2\), 0, or 2?

Graph of f

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At −2, f is not continuous.

At 0, the graph is a straight piece of slope \(-1\), so the derivative exists.

At 2, the point is filled and the slopes change from \(-1\) to 1.

Answer: \(2\)

Use the graph. At which of \(-2\), 0, and 2 is f not continuous?

Graph of f

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At −2 the limit is 4 and the function value is 1.

At 0 and at 2, the graph meets itself.

Answer: \(-2\)

For \(f(x) = (x - 5)^{1/3}\), the difference quotient at 5 simplifies to \(h^{-2/3}\). Does \(f'(5)\) exist?

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As \(h \to 0\), \(h^{-2/3}\) grows without a bound.

The limit is not a real number.

f is not differentiable at 5. The tangent is vertical.

Answer: no

f is differentiable at \(x = 1\) and \(f'(1) = 4\). Which must be true?

A. \(\lim_{x \to 1} f(x) = f(1)\)

B. \(f(1) = 4\)

C. f has a corner at \(x = 1\)

D. \(f'(x) = 4\) for every x

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Differentiability implies continuity, so the limit equals \(f(1)\).

\(f'(1) = 4\) is a slope, not the function value.

A corner would mean the derivative does not exist.

Answer: A. the limit equals f(1)

Level 3

Hard

Make a piecewise function differentiable, or show that no constant works.

Example

Let \(f(x) = ax^2 + 5\) for \(x \leq 1\), and \(f(x) = 4x + b\) for \(x \gt 1\). Find a and b so that f is differentiable at \(x = 1\).

Differentiability requires continuity, so the two formulas must agree at \(x = 1\).

\(a + 5 = 4 + b\)

Left-hand derivative: \(\dfrac{d}{dx}(ax^2 + 5) = 2ax\), so at 1 the slope is \(2a\).

Right-hand derivative: \(4\).

\(2a = 4\)

\(a = 2\)

Substitute into the value equation.

\(2 + 5 = 4 + b\)

\(b = 3\)

Answer: \(a = 2\), \(b = 3\)

Let \(f(x) = 3x + k\) for \(x \lt 2\), and \(f(x) = x^2 + 1\) for \(x \geq 2\). Is there a value of k that makes f differentiable at 2?

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Continuity at 2:

\(3(2) + k = 2^2 + 1\)

\(6 + k = 5\)

\(k = -1\)

Left-hand derivative: 3.

Right-hand derivative: \(2x\) at 2 is 4.

\(3 \neq 4\)

The slopes cannot match. No value of k makes f differentiable at 2.

With \(k = -1\), f is continuous and has a corner.

Answer: no

Let \(f(x) = x^3 - 2\) for \(x \lt 1\), and \(f(x) = mx + n\) for \(x \geq 1\). Find m and n so that f is differentiable at 1.

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Left-hand value at 1:

\(1 - 2 = -1\)

Right-hand value at 1:

\(m + n\)

\(m + n = -1\)

Left-hand derivative: \(3x^2\) at 1 is 3.

Right-hand derivative: \(m\).

\(m = 3\)

\(3 + n = -1\)

\(n = -4\)

Answer: \(m = 3\), \(n = -4\)

Let \(f(x) = x^2 + 1\) for \(x \leq 2\), and \(f(x) = 4x + k\) for \(x \gt 2\). Find k so that f is differentiable at 2.

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Left-hand derivative at 2: \(2x = 4\).

Right-hand derivative: 4.

The slopes already agree.

Continuity:

\(2^2 + 1 = 4(2) + k\)

\(5 = 8 + k\)

\(k = -3\)

Answer: \(k = -3\)

Explain why \(f(x) = x^{1/3}\) is continuous at 0 but not differentiable at 0.

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\(f(0) = 0\)

\(\lim_{x \to 0} x^{1/3} = 0\)

f is continuous at 0.

The difference quotient is

\(\dfrac{h^{1/3} - 0}{h} = h^{-2/3}\)

As \(h \to 0\), \(h^{-2/3}\) is unbounded.

The derivative is not a real number.

Answer: continuous, not differentiable

f is differentiable at \(x = 6\). A student checks only that the two pieces of a formula give the same y-value at 6 and stops. What else has to be checked?

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Equal y-values give continuity.

Differentiability also needs the left-hand derivative to equal the right-hand derivative.

If those slopes differ, the graph has a corner.

Answer: the one-sided derivatives must be equal

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