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Calculus BC · Unit 2 · 2.4
AP Calculus · Differentiation: Definition and Fundamental Properties
Connecting Differentiability and Continuity
Four ways a derivative fails — and three of them happen at perfectly continuous points.
key idea$\text{Differentiable}\Rightarrow\text{continuous. The converse is FALSE.}$
The categories
A derivative is a limit of slopes, so it can fail for several different reasons. Naming the type of failure is what the exam asks for — and only the last of these four also breaks continuity.
Four ways $f'(0)$ can fail. The first three functions are perfectly continuous at $0$; only the fourth is not. Continuity is necessary for differentiability but nowhere near sufficient.| Type | What the one-sided slopes do | Example | Continuous there? |
|---|
| Corner | both finite, but different | $\lvert x\rvert$ at $0$: $-1$ and $+1$ | yes |
| Cusp | $-\infty$ on one side, $+\infty$ on the other | $x^{2/3}$ at $0$ | yes |
| Vertical tangent | both $\to+\infty$ (or both $-\infty$) | $\sqrt[3]{x}$ at $0$ | yes |
| Discontinuity | no slope limit at all | a jump at $0$ | no |
The right-hand column is the point of the whole topic: three of the four failures happen at points where the function is continuous.
Telling them apart
the decision order
- Check continuity first — if it fails, differentiability fails and you are done.
- If continuous, compare the one-sided slopes.
- Both finite and unequal ⇒ corner.
- Opposite infinities ⇒ cusp; same infinity ⇒ vertical tangent.
| Statement | True? | Why |
|---|
| differentiable at $c$ ⇒ continuous at $c$ | true | the defining limit cannot exist across a break |
| continuous at $c$ ⇒ differentiable at $c$ | false | $\lvert x\rvert$ at $0$ — continuous, no derivative |
| not continuous at $c$ ⇒ not differentiable at $c$ | true | the contrapositive of the first row |
| not differentiable at $c$ ⇒ not continuous at $c$ | false | the first three figure panels are all counterexamples |
Rows 1 and 3 say the same thing; rows 2 and 4 say the same thing. Only one direction of the implication is valid.
Worked examples
EX 1
Classify a corner
$\lvert x-2\rvert$Is $f(x)=\lvert x-2\rvert$ differentiable at $x=2$?
1
Check continuity
$\lim_{x\to2}\lvert x-2\rvert=0=f(2)$, so $f$ is continuous at $2$.
2
Left-hand slope
For $x\lt 2$, $f(x)=2-x$ with slope $-1$.
3
Right-hand slope
For $x\gt 2$, $f(x)=x-2$ with slope $+1$.
4
Compare
Both are finite but $-1\neq+1$, so the defining limit does not exist.
5
Classify
Not differentiable at $x=2$ — a corner. Continuous but not differentiable.
EX 2
Cusp versus vertical tangent
$x^{2/3}$ and $\sqrt[3]{x}$Both fail at $x=0$. What distinguishes them?
1
Differentiate each
$\dfrac{d}{dx}\left[x^{2/3}\right]=\dfrac{2}{3x^{1/3}}$ and $\dfrac{d}{dx}\left[x^{1/3}\right]=\dfrac{1}{3x^{2/3}}$.
2
Slopes near $0$ for $x^{2/3}$
The cube root in the denominator is negative for $x\lt 0$ and positive for $x\gt 0$: slopes $\to-\infty$ then $+\infty$.
3
Slopes near $0$ for $x^{1/3}$
$x^{2/3}$ is a square of a real, so always positive: slopes $\to+\infty$ from both sides.
4
Classify
$x^{2/3}$ has a cusp (the two branches meet at a point); $\sqrt[3]{x}$ has a vertical tangent (the curve passes smoothly through, just vertically).
5
The discriminator
Opposite infinities give a cusp; matching infinities give a vertical tangent. Both are continuous and neither is differentiable.
EX 3
Use the implication
reasoning$g$ is differentiable at $x=5$. What can you conclude about $\lim\limits_{x\to5}g(x)$?
1
Apply the theorem
Differentiability at a point implies continuity there.
2
Unpack continuity
So $\lim_{x\to5}g(x)$ exists, $g(5)$ exists, and they are equal.
3
State the answer
$\lim_{x\to5}g(x)=g(5)$.
4
Note what you cannot conclude
Nothing about $g'$ being continuous, and nothing about behaviour anywhere else. The implication is pointwise.
EX 4
Make a piecewise function differentiable
solveFind $a$ and $b$ so that $h(x)=\begin{cases}x^{2},& x\le1\\ ax+b,& x\gt 1\end{cases}$ is differentiable at $x=1$.
1
Continuity first
Differentiability requires continuity, so match the values: $1^{2}=a(1)+b$, giving $a+b=1$.
2
Match the slopes
Left derivative: $\dfrac{d}{dx}\left[x^{2}\right]=2x\to2$ at $x=1$. Right derivative: $a$.
4
Solve for $b$
From $a+b=1$: $b=-1$.
5
Why two conditions
Continuity gives one equation, slope-matching the second. Slope-matching alone is not enough — the pieces could be parallel but not touching.
Watch out
The implication runs one way only
Differentiable ⇒ continuous. The converse is false, and the first three figure panels are the standard counterexamples.
Matching slopes does not give differentiability
In Example 4, $a=2$ with $b=7$ would give parallel pieces that do not meet. You need both the value and the slope to match.
Cusp and vertical tangent are different answers
Both have infinite one-sided slopes, but a cusp has them pointing opposite ways. If a question asks you to classify, the distinction is the mark.
“Smooth” is not a synonym for continuous
$\lvert x\rvert$ is continuous everywhere and smooth nowhere at $0$. Use the precise word.