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Monotonicity and constancy from the sign of the derivative, antiderivatives up to a constant, and L'Hopital's rule with its hypotheses.
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 deduce a function's monotonicity and constancy from the sign of its derivative, prove that two functions with the same derivative on an interval differ by a constant, and apply L'Hopital's rule only where its hypotheses hold. You will also be able to say why every one of these results needs the domain to be a single interval, and what goes wrong when it is not.
The mean value theorem, in the form: the change in a function across an interval is the derivative somewhere times the width. Every result in this lesson is that identity read with a different hypothesis on the derivative.
A critical point is one where the derivative vanishes. A quotient is an indeterminate form at a point when both parts tend to zero, or both grow without bound — which is the hypothesis L'Hopital's rule requires, and the only situation in which the rule says anything.
For $x < y$ in an interval where $f$ is continuous and differentiable, $$f(y) - f(x) = f'(p)\,(y - x) \quad \text{for some } p \in (x, y).$$ The factor $y - x$ is positive, so the sign of the change is the sign of $f'(p)$. Everything follows.
Monotonicity. $f' > 0$ on an interval gives $f$ strictly increasing there; $f' < 0$ gives strictly decreasing; $f' \ge 0$ gives non-decreasing.
Constancy. $f' = 0$ on an interval gives $f$ constant there.
Uniqueness of antiderivatives. $f' = g'$ on an interval gives $f - g$ constant, by applying constancy to the difference.
L'Hopital's rule. If $f(c) = g(c) = 0$, both are differentiable near $c$, $g' \ne 0$ near $c$, and $f'/g'$ has a limit, then $f/g$ has the same limit. It is proved from the Cauchy mean value theorem, which is the same tilt applied to a pair of functions at once.
Every one needs an interval. The point $p$ has to lie between $x$ and $y$ and inside the domain. On a domain in two pieces the results hold piece by piece and not across the gap, and $\operatorname{sgn}$ is the example that shows it.
Another way: picture
Take any two points of the graph and draw the chord between them. The theorem says some tangent is parallel to it. If every tangent slopes upward, so does every chord — and a function all of whose chords slope upward is exactly an increasing one.
Another way: steps
To describe a function's shape from its derivative:
For a limit of indeterminate form: check the form, differentiate top and bottom separately, take the limit again, and repeat only while the form stays indeterminate.
The rule is stated for $0/0$ and for $\infty/\infty$, and it requires:
Only then does $\lim f/g = \lim f'/g'$.
Why it is true, in the simplest case. If $f(c) = g(c) = 0$ and both are differentiable at $c$ with $g'(c) \ne 0$, $$\frac{f(x)}{g(x)} = \frac{(f(x)-f(c))/(x-c)}{(g(x)-g(c))/(x-c)} \longrightarrow \frac{f'(c)}{g'(c)}.$$ No new theorem is needed for that version; the general one needs the Cauchy mean value theorem.
Two failures worth knowing. Applied to a determinate form it gives a wrong answer silently. And the converse fails: $\lim f'/g'$ may not exist while $\lim f/g$ does — $\frac{x + \sin x}{x}$ at infinity tends to $1$, while the quotient of derivatives oscillates for ever. So a rule that gives nothing has not shown the original limit does not exist.
| Derivative on an interval | Function there |
|---|---|
| $f' > 0$ | strictly increasing |
| $f' \ge 0$ | non-decreasing |
| $f' = 0$ | constant |
| $f'$ increasing | convex; chords above the graph |
| $f'(c) = 0$ and $f''(c) > 0$ | a local minimum at $c$ |
The fourth and fifth rows are the second-derivative results, and they are the first rows applied to $f'$ rather than to $f$. That is the pattern worth carrying forward: a statement about $f''$ is a statement about the shape of $f'$, and the shape of $f'$ governs the shape of $f$.
One caution about the first row: strictly increasing does not give $f' > 0$. The function $x^3$ is strictly increasing and its derivative vanishes at $0$. The implication runs one way.
Applying them off an interval. A domain with a point removed allows a different constant on each side, and $\operatorname{sgn}$ has zero derivative everywhere it is defined.
Reading monotonicity backwards. Strictly increasing does not require a strictly positive derivative; $x^3$ is the standing counterexample.
Using L'Hopital on a determinate form. The hypothesis is that the form is indeterminate. Without it the rule returns a number, and the number is wrong.
Treating a failed L'Hopital as a verdict. If the quotient of derivatives has no limit, nothing follows about the original quotient, which may converge perfectly well.
Let $f(x) = x^3 - 3x$. Then $f'(x) = 3(x-1)(x+1)$, vanishing at $\pm 1$.
Critical points first.
$f' > 0$ below $-1$ and above $1$; $f' < 0$ between them.
Sign on each interval between the critical points.
So $f$ increases on each outer interval and decreases between, with a local maximum at $-1$ and a local minimum at $1$.
Each statement about one interval only.
$\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}$: at $0$ both parts vanish, so the form is $0/0$.
Check the form before applying.
Differentiating gives $\dfrac{\sin x}{2x}$, which is again $0/0$, so the rule may be applied once more.
Check the form again, not just once.
Differentiating again gives $\dfrac{\cos x}{2} \to \dfrac{1}{2}$. Each application needed its own check.
The hypothesis is checked every time.
Differentiate: the two derivatives are $\dfrac{1}{1+x^2}$ and $-\dfrac{1}{1+x^2}$, which sum to zero.
Differentiate the whole expression.
The positive reals form an interval, so the corollary applies and the sum is constant there.
Check that the domain is one interval.
Evaluating at $x = 1$ gives $\pi/4 + \pi/4 = \pi/2$, so the constant is $\pi/2$. On the whole domain of both functions the sum takes a different constant on the negative side, which is the domain caution in action rather than a curiosity.
On an interval, match each hypothesis about the derivative to what follows about the function.
| the function strictly increases | the function strictly decreases | the function is constant | the two differ by a constant | |
|---|---|---|---|---|
| the derivative is positive throughout | ||||
| the derivative is negative throughout | ||||
| the derivative is zero throughout | ||||
| two functions have the same derivative throughout |
Evaluate $\displaystyle\lim_{x \to 0} \dfrac{5x + x^2}{7x - x^3}$.
Answer:
Let $f(x) = x^3 - 31^2 x$. Give the set of $x$ at which $f'(x) > 0$.
This task has no paper form; do it on a device.
Put the steps of the proof that a function with zero derivative on the interval from $0$ to $4$ is constant into order.
Number the steps in order (write the number in the box):
Here is a computation of $\displaystyle\lim_{x \to 0} \dfrac{x + 3}{2x + 1}$. Mark the one step that is not permitted.
This task has no paper form; do it on a device.
Let $f(x) = 5$ for $x > 0$ and $f(x) = -5$ for $x < 0$, defined on the reals with $0$ removed. Its derivative is zero everywhere on its domain, and it is not constant. What has gone wrong?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $f(x) = x^3 - 34^2 x$. Give the set of $x$ at which $f'(x) > 0$.
This task has no paper form; do it on a device.
You can read a function's shape off the sign of its derivative, prove the constancy corollary, and apply L'Hopital's rule after checking the form. Say in your own words why a function with zero derivative on a domain in two pieces need not be constant. Next: approximating a function by a polynomial, with a remainder that is bounded rather than ignored.
10. Your turn: show that $\arctan x + \operatorname{arccot} x$ is constant on the positive reals, step 3