A. Rate of Change · Limits and Rates of Change

When Does a Function Have No Tangent? Differentiability

A1.2 A1.5About 70 minutes

Learning goals

I can decide whether a function is differentiable at a point by comparing one-sided limits of the difference quotient.
I can identify corners, cusps, vertical tangents and discontinuities as points where a function is not differentiable.
I can explain why a differentiable function must be continuous, and why the reverse is not true.
I can choose constants so that a piecewise function is smooth (continuous with no corner).

Words to know

differentiable at a
f′(a) exists: the limit of the difference quotient is a real number.

one-sided difference quotient limits
lim (h → 0⁻) and lim (h → 0⁺) of (f(a + h) − f(a))/h: the slopes approached by secants from the left and from the right.

corner
A point where the one-sided limits of the difference quotient are different real numbers.

cusp
A sharp point where the difference quotient approaches ∞ on one side and −∞ on the other.

vertical tangent
A point where the difference quotient approaches ∞ (or −∞) from both sides, so the tangent is vertical.

smooth join
Two pieces of a function that meet with no break and with equal slopes at the joining point.

Back to the start, with better tools

In the lesson Secants, Tangents and Smooth Functions you saw pictures of graphs with no tangent at a point: corners, cusps, breaks and vertical tangents. The curriculum's sample problem asked why the instantaneous rate of change cannot be found at such a point. Now you can answer with limits.

f is differentiable at a exactly when lim (h → 0) (f(a + h) − f(a))/h exists as a real number.

Like any two-sided limit, this one exists only if the left limit (h → 0⁻, secants from the left) and the right limit (h → 0⁺, secants from the right) exist and are equal.

Worked example

Example 1: A corner

Is f(x) = |x − 3| differentiable at x = 3? Is it continuous there?

f(3) = 0 and f(3 + h) = |h|. The difference quotient is |h|/h.
Right (h > 0): |h|/h = h/h = 1, so the right limit is 1.
Left (h < 0): |h|/h = −h/h = −1, so the left limit is −1.
The one-sided limits differ: the limit of the difference quotient does not exist.
Continuity: lim (x → 3) |x − 3| = 0 = f(3), so f is continuous at 3.

f is continuous but not differentiable at x = 3. The graph has a corner there.

The difference quotient q(h) as a function of h. For y = x² at a = 1 it is the line 2 + h, which approaches 2 from both sides. For y = |x| at a = 0 it jumps from −1 to 1, so it has no limit.

Worked example

Example 2: Cusps and vertical tangents

Investigate the difference quotients at x = 0 for f(x) = ∛(x²) and g(x) = ∛x.

For f: f(h) − f(0) = ∛(h²), so the quotient is ∛(h²)/h = 1/∛h. With h = 0.001: 1/0.1 = 10; with h = 0.000 001: 100. With h = −0.001: −10; with h = −0.000 001: −100.
So the quotient → ∞ from the right and → −∞ from the left: a cusp.
For g: the quotient is ∛h/h = 1/∛(h²). With h = ±0.001: 1/0.01 = 100; with h = ±0.000 001: 10 000. It → ∞ from both sides: a vertical tangent.
In both cases the limit is not a real number.

Neither function is differentiable at 0, although both are continuous there. f has a cusp; g has a vertical tangent (the y-axis).

Differentiable means continuous

If f is differentiable at a, then f is continuous at a. Here is why. For h ≠ 0,

f(a + h) − f(a) = [(f(a + h) − f(a))/h] × h.

As h → 0, the bracket approaches f′(a) (a real number) and h approaches 0. By the product law, the right side approaches f′(a) × 0 = 0. So f(a + h) → f(a), which is continuity.

So a discontinuity (hole, jump or asymptote) always destroys differentiability. But the reverse fails: |x|, ∛(x²) and ∛x are continuous at 0 and still not differentiable. The curriculum's word smooth (continuous with no corners) is meant to rule out all of these trouble spots: on a smooth stretch of graph, every point has a tangent with a real-number slope.

at x = a	continuous?	differentiable?
smooth point	yes	yes
corner, cusp, vertical tangent	yes	no
hole, jump, asymptote	no	no

Differentiable always implies continuous, but not the other way around.

Worked example

Example 3: Does a piecewise function join smoothly?

Decide whether each function is differentiable at x = 1.

(a) f(x) = x² for x ≤ 1, and f(x) = 2x − 1 for x > 1 (b) g(x) = x² for x ≤ 1, and g(x) = 3x − 2 for x > 1

(a) Continuity: both pieces give 1 at x = 1 ✓.
Left quotient (h < 0, use x²): ((1 + h)² − 1)/h = 2 + h → 2. Right quotient (h > 0, use the line): (2(1 + h) − 1 − 1)/h = 2h/h = 2.
Both sides give 2, so f is differentiable at 1 and f′(1) = 2.
(b) Continuity: 1² = 1 and 3(1) − 2 = 1 ✓. Left quotient → 2 (as before). Right quotient: (3(1 + h) − 2 − 1)/h = 3.
The sides disagree (2 and 3): a corner.

(a) Differentiable, f′(1) = 2 (the line is the tangent to the parabola). (b) Continuous but not differentiable: a corner at (1, 1).

Worked example

Example 4: Designing a smooth ramp

An engineer joins a flat road (y = 0 for x ≤ 0) to a straight ramp with slope 0.1 using a curved section y = kx² for 0 ≤ x ≤ 20 (x and y in metres). Find k so the join at x = 20 is smooth, and write the ramp's equation.

At x = 0: the curve y = kx² has slope 2k(0) = 0, which matches the flat road, and both have height 0. ✓
At x = 20: the slope of kx² at x = a is 2ka (first principles: (k(a + h)² − ka²)/h = 2ka + kh → 2ka). So the slope at 20 is 40k.
Smooth join: 40k = 0.1, so k = 0.0025.
Height at x = 20: 0.0025 × 400 = 1 m. The ramp passes through (20, 1) with slope 0.1: y = 1 + 0.1(x − 20) = 0.1x − 1.

k = 0.0025, and the ramp is y = 0.1x − 1 for x ≥ 20. With no corners, a car's direction changes gradually instead of suddenly.

A road profile built from three pieces: flat, curved (dashed) and straight. At each join the heights and the slopes match, so the whole profile is smooth.

Just for fun

Engineers really do avoid corners in roads and railways. Transition curves ease drivers and passengers from straight sections into bends so nobody gets an instant jolt. A corner in a function is just a jolt in a graph.

Hands-on activity

Corner or curve? A paper road

You need: grid paper, a pencil, a ruler, a toy car (or a coin), and some tape.

On grid paper, draw a road that is flat, then turns sharply upward at a corner (like the letter L leaned back).
Draw a second road that is flat, then curves gradually into the same upward slope.
Slowly push the toy car along each road. Feel and watch what happens at the join.
For each road, find the slope just before and just after the join by counting grid squares on a short secant on each side.
Explain which road is differentiable at the join, using the words left slope and right slope.

What to look for: on the corner road, the left slope is 0 and the right slope is not 0, so the car's direction changes all at once (it bumps or jumps). On the curved road, the left and right slopes both approach the same value at each join, so the car turns smoothly. Only the curved road is differentiable at the join.

Your turn

Is f(x) = |x + 2| differentiable at x = −2? Show the one-sided quotients.

Is f(x) = |x + 2| continuous at x = −2?

Is f(x) = x|x| differentiable at x = 0? If so, find f′(0).

Is f(x) = ∛(x − 1) differentiable at x = 1? What does the graph look like there?

f(x) = x² for x < 2 and f(x) = 4x − 4 for x ≥ 2. Is f differentiable at 2?

Can the parking-fee step function from Limit Laws and Continuity be differentiable at t = 2 h? Explain.

True or false: if f is continuous at a, then f is differentiable at a.

More practice: warm-up

What does "f is differentiable at a" mean?

Give a function with a corner at x = 0.

If f is differentiable at a, must it be continuous there?

Is f(x) = x² differentiable at x = 0? What is f′(0)?

Secants at a point approach a vertical line. What is this called?

More practice: on your own

Find the one-sided limits of the difference quotient of f(x) = |x² − 4| at x = 2. Is f differentiable there?

f(x) = x² + 1 for x ≤ 1 and f(x) = ax + b for x > 1. Find a and b so that f is differentiable at 1.

Give an example of a function that is continuous at x = 2 but not differentiable there, and explain.

Is f(x) = √|x| differentiable at 0? Describe the point.

A ball dropped from 4.9 m has height 4.9 − 4.9t² for 0 ≤ t ≤ 1 and, after bouncing, 7(t − 1) − 4.9(t − 1)² for t ≥ 1. Find the one-sided velocities at t = 1. Is the height differentiable there?

Challenge

Write out the proof that differentiability at a implies continuity at a.

A curve y = kx² starts at a flat road at x = 0 and must join a straight ramp of slope 0.08 smoothly at x = 25. Find k and the ramp's equation.

Where is f(x) = |x − 1| + |x − 3| not differentiable? Find the slope on each piece.

Check your understanding

Quiz

f(x) = |x − 5|. What are the one-sided limits of the difference quotient at x = 5?

Which statement is true?

For f(x) = ∛x, the difference quotient at 0 approaches ∞ from both sides. What does the graph have at the origin?

f(x) = x² for x ≤ 2 and f(x) = mx + b for x > 2 is differentiable at 2. What is m?

Is f(x) = x|x| differentiable at x = 0?

A ball's height graph has a sharp corner at the moment of a bounce. What does this say about the model?

For parents and teachers

Big idea (A1.2, A1.5): f′(a) exists only when the difference quotient has the same real-number limit from both sides. Corners, cusps, vertical tangents and discontinuities all break this. Differentiable functions are always continuous, but continuous functions can still have corners. This answers the curriculum's sample question about why no instantaneous rate exists at a point like P.

How to help: Do the paper-road activity together. Ask your teen to explain to you, using "left slope" and "right slope", why one road is smooth and the other is not.

Common difficulties:

Checking only continuity and stopping there. Continuity is necessary but not enough.
Using the wrong piece of a piecewise function on each side (left quotient uses the left piece).
Getting the sign of |h| wrong for h < 0 (|h| = −h).
Mixing up cusps (opposite infinities) and vertical tangents (same infinity).