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Differentiating with the point left unnamed, the implication to continuity, and the corner, cusp and vertical tangent.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to differentiate with the point left as a variable to get the derivative function and its own domain, prove and use the fact that differentiability implies continuity while knowing the standard counterexample to the converse, decide differentiability at a junction by checking continuity first and then matching the one-sided derivatives, classify a failure as a corner, a cusp or a vertical tangent, and choose constants that make a piecewise function join smoothly.
Leave the point unnamed and the calculation of the last lesson produces a function: $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h},$$ defined at every $x$ where that limit exists. The domain of $f'$ can be smaller than the domain of $f$, and the points it loses are the interesting ones.
Differentiable implies continuous. If $f'(a)$ exists then $$f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a) \longrightarrow f'(a) \cdot 0 = 0,$$ so $f(x) \to f(a)$. Short, and worth following once: the product law does all of it.
The converse is false, and the counterexample is the one function everybody already knows. $|x|$ is continuous at $0$ and has no derivative there. So differentiability is a strictly stronger condition, and "the graph has no break" does not buy "the graph has a tangent".
The three ways it fails at a point where $f$ is continuous:
| Failure | The one-sided quotients | Example at $0$ |
|---|---|---|
| corner | two different finite limits | $\|x\|$ |
| cusp | run to $+\infty$ and $-\infty$ | $x^{2/3}$ |
| vertical tangent | both run the same way to $\pm\infty$ | $x^{1/3}$ |
And the fourth way: $f$ is not continuous at all, in which case the numerator does not vanish, the quotient is a non-zero number over something tending to $0$, and no derivative can exist.
Another way: picture
Zoom in on a graph at a point. A differentiable function looks straighter the closer you look, and in the limit is indistinguishable from its tangent — that is what differentiable means, made visual. $|x|$ at the origin does not: the corner stays a corner at every magnification, which is why no line can be the tangent there.
Another way: steps
To decide differentiability at $a$:
A piecewise function with a junction at $a$ raises two separate questions, and they have to be answered in order.
Continuous? One equation: the two pieces agree at $a$.
Differentiable? A second equation on top of it: the two derivatives agree at $a$. And the order matters — without continuity the numerator of the difference quotient does not vanish, so the quotient runs to infinity and differentiability cannot be rescued by any amount of slope-matching.
So a smooth join costs two conditions, and a piecewise definition with two free constants has exactly enough freedom to meet them. The line that results is the tangent line to the other piece — which is a good way to remember it, and a first glimpse of why a curve and its tangent agree "to first order".
A warning about matching the pieces' derivative formulas. What differentiability actually requires is that the two one-sided derivatives of $f$ agree — limits of difference quotients of $f$ itself. For the familiar pieces those coincide with the limits of the pieces' derivative formulas, and that is what makes the quick method valid. It is not a definition, and for a piece like $x^2\sin(1/x)$, whose derivative formula has no limit at $0$ although the derivative exists there, the quick method gives the wrong answer.
Reading "continuous" as "differentiable". The implication runs one way. $|x|$ at $0$, $|x-3|$ at $3$, and any function with a corner are all continuous and not differentiable.
Checking slopes before checking continuity. If the pieces do not meet, matching their slopes proves nothing.
Thinking a formula that is defined everywhere has a derivative everywhere. $x^{1/3}$ is defined on all of $\mathbb{R}$ and has no derivative at $0$.
Forgetting that $f'$ has its own domain. "$f'(x) = 2x$ for all $x$" is a claim about where the limit exists, not only about the algebra that produced the formula.
Assuming $f'$ is continuous. It need not be. $f(x) = x^2\sin(1/x)$ with $f(0) = 0$ is differentiable everywhere, and $f'$ has a genuine discontinuity at $0$. Differentiability of $f$ says nothing about continuity of $f'$, and the functions where it does have a name of their own — $C^1$.
The commonest wrong method for a piecewise function is to differentiate each piece, evaluate both formulas at the junction, and declare the function differentiable when the two numbers agree. It usually gives the right answer, and it is not the definition — which is about the limit of the difference quotient of $f$, not about the pieces' formulas.
The gap shows in both directions. $f(x) = x^2\sin(1/x)$ with $f(0) = 0$ is differentiable at $0$, with $f'(0) = 0$ by the squeeze theorem, and yet $f'(x) = 2x\sin(1/x) - \cos(1/x)$ has no limit at $0$ at all — the quick method would find no value to match and wrongly conclude failure. And skipping the continuity check lets the quick method declare a function with a jump differentiable, because two pieces can have equal slopes and still not meet.
Use the quick method, and know what it is standing on.
$f(x) = |x - 1| + |x - 4|$. Each absolute value is continuous, so the sum is.
Continuity is easy.
On $x < 1$ it is $5 - 2x$ with slope $-2$; on $1 < x < 4$ it is $3$ with slope $0$; on $x > 4$ it is $2x - 5$ with slope $2$.
Three pieces, three slopes.
At $x = 1$ the one-sided slopes are $-2$ and $0$; at $x = 4$ they are $0$ and $2$. Two corners. The domain of $f$ is $\mathbb{R}$ and the domain of $f'$ is $\mathbb{R}$ with two points removed.
$f(x) = x^2$ for $x \le 1$ and $f(x) = kx + c$ for $x > 1$. Derivative match: $2x$ at $x = 1$ is $2$, so $k = 2$.
Slopes first, since $k$ appears only there.
Value match: $1 = 2(1) + c$, so $c = -1$.
Then continuity fixes $c$.
The joining line is $y = 2x - 1$, which is precisely the tangent to $y = x^2$ at $(1,1)$. It could not have been anything else: same point, same slope.
The absolute value bends where its inside changes sign: $x^2 - 9 = 0$ at $x = \pm 3$.
Find the sign changes.
At those points the inside crosses zero with non-zero slope ($2x = \pm 6$), so the graph reflects and a corner forms.
Two points, $x = 3$ and $x = -3$. Everywhere else $f$ is a polynomial locally and is differentiable.
Match each one-sided slope behavior to its derivative conclusion.
| Derivative exists and equals 4 | Corner: no derivative | Cusp: no finite derivative | Vertical tangent: no finite derivative | |
|---|---|---|---|---|
| Left and right slopes both approach 4 | ||||
| Left slope approaches -1; right slope approaches 1 | ||||
| Slopes approach $-\infty$ and $+\infty$ | ||||
| Both slopes approach $+\infty$ |
Match each local feature to its differentiability conclusion.
| Not differentiable because it is not continuous | Continuous but not differentiable | Differentiable | No finite derivative at that point | |
|---|---|---|---|---|
| A jump discontinuity | ||||
| $f(x)=|x|$ at 0 | ||||
| A polynomial at any real input | ||||
| A vertical tangent |
Match each local behavior at a point to the differentiability conclusion.
| Differentiable with derivative 4 | A corner: continuous but not differentiable | A cusp: no finite derivative | Discontinuous, therefore not differentiable | |
|---|---|---|---|---|
| Both one-sided slopes approach 4 | ||||
| Left slope approaches -1; right slope approaches 1 | ||||
| Left slope approaches $-\infty$; right slope approaches $+\infty$ | ||||
| The function has different left and right values |
$f(x) = 3x^2 + 6x$. Compute $f'(x)$ from the definition, leaving $x$ as a variable.
Answer:
Is $f(x) = x^{1/3}$ differentiable at $x = 0$?
$f(x) = 5x^2$ for $x \le 3$ and $f(x) = kx + c$ for $x > 3$. With $k$ chosen to make $f$ differentiable at $3$, what is $c$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f(x) = x^2$ for $x \le 5$ and $f(x) = kx + c$ for $x > 5$. For $f$ to be differentiable at $5$, what must $k$ be?
Answer:
You can find a derivative function, decide differentiability at a junction, and name the three failures. Say in your own words why continuity has to be checked before slopes are matched. Next: the rules, which replace the definition for every function you will meet.
8. Your turn: where does $f(x) = |x^2 - 9|$ fail to be differentiable?, step 3