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Fermat's theorem with its hypotheses, the three places an extremum can be, and the method that cannot miss.
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 state Fermat's theorem with both hypotheses and name the counterexample each one excludes, find every critical point including those where the derivative fails to exist, classify a stationary point by the first or second derivative test and recognise when the second is silent, apply the closed-interval method with the endpoints in the candidate list, and say why the extreme value theorem is what turns that list into a proof rather than a search.
$f$ has a local maximum at $c$ if $f(c) \ge f(x)$ for all $x$ near $c$, and a global (or absolute) maximum if the inequality holds everywhere on the domain. Minima mirror it.
Fermat's theorem. If $f$ has a local extremum at an interior point $c$ and $f'(c)$ exists, then $f'(c) = 0$.
Read the hypotheses, because each excludes a case that really happens:
So the places an extremum can hide are exactly three: where $f' = 0$, where $f'$ does not exist, and at an endpoint. The first two are the critical points.
Necessary, not sufficient. $f'(c) = 0$ does not make $c$ an extremum: $x^3$ at $0$ has a horizontal tangent and passes straight through.
The closed-interval method. For $f$ continuous on $[a, b]$:
It works because the extreme value theorem has already promised that both exist. Without that guarantee the method would be a search with no assurance of finding anything.
Another way: picture
A hilly path between two fence posts. The high point is either at the top of a hill, where the path is momentarily level, or at a sharp peak where it has no well-defined slope, or at one of the posts. There is nowhere else it can be — and that list of three is the whole method.
Another way: steps
Two tests, and they answer the same question with different amounts of information.
First derivative test. Look at the sign of $f'$ either side of $c$:
| Left | Right | At $c$ |
|---|---|---|
| $+$ | $-$ | local maximum |
| $-$ | $+$ | local minimum |
| same sign | same sign | neither |
Second derivative test. If $f'(c) = 0$: $f''(c) > 0$ gives a minimum, $f''(c) < 0$ a maximum, and $f''(c) = 0$ gives no information.
The second test is quicker where it works, and where $f''(c) = 0$ it is genuinely silent rather than suggestive. All three of $x^4$ (minimum), $-x^4$ (maximum) and $x^3$ (neither) have $f'(0) = f''(0) = 0$, so the same evidence is consistent with every answer. In that case the first derivative test decides it, and always can.
Only solving $f' = 0$. Points where $f'$ does not exist are critical too, and for $|x|$ that is the only candidate there is.
Forgetting the endpoints. On a closed interval they are always candidates, and often the answer. This is the most common way a correct calculation reaches a wrong conclusion.
Reading a stationary point as an extremum. Necessary is not sufficient; $x^3$ is the standing counterexample.
Comparing derivatives instead of values. Having found the candidates, the comparison is between $f$ at them. The derivative has done its job and has nothing further to say.
Using the second derivative test when it is silent. $f''(c) = 0$ means "no conclusion", not "inflection".
Applying the closed-interval method to an open or unbounded interval. The extreme value theorem does not apply, so nothing guarantees a maximum exists; the method can then return a largest candidate that is not one.
Optimisation gets compressed in memory to solving $f' = 0$, and that loses two of the three places an answer can be. A maximum at an endpoint has a non-zero derivative; a maximum at a corner has none. Neither is exotic — endpoints turn up in every constrained problem, and corners in every absolute value.
The honest version of the method is a list of candidates, assembled from all three sources, followed by a comparison of values. That is slower to say and it cannot miss, which is the trade the method is making.
And behind it stands the extreme value theorem. It is what turns the list into a proof: the maximum exists, it must be one of these, therefore the largest of these is it. Drop the closed interval and the reasoning collapses — which is the connection between lesson 10's hypotheses and this lesson's procedure, and it is not decoration.
$f(x) = x^{2/3}(x - 5)$ on $[-1, 6]$. Expanding: $f = x^{5/3} - 5x^{2/3}$, so $f' = \tfrac53 x^{2/3} - \tfrac{10}{3}x^{-1/3}$.
A fractional power hints at a second kind of critical point.
$f' = 0$: multiply by $3x^{1/3}$ to get $5x - 10 = 0$, so $x = 2$. And $f'$ does not exist at $x = 0$, because of the $x^{-1/3}$.
Two critical points, found two different ways.
Candidates $-1, 0, 2, 6$. Values: $-6$, $0$, $\approx -4.76$, $\approx 3.30$. Maximum $\approx 3.30$ at the right endpoint; minimum $-6$ at the left. Neither extreme is at a critical point — and both would have been missed by a search that skipped the ends.
$f(x) = x^4$. $f'(x) = 4x^3$, zero at $x = 0$; $f''(x) = 12x^2$, also zero at $0$. The second derivative test concludes nothing.
Silent, not suggestive.
First derivative test: $f' < 0$ for $x < 0$ and $f' > 0$ for $x > 0$.
Decreasing then increasing.
So $x = 0$ is a local minimum — and a global one. The first derivative test decided what the second could not, and it always can.
$f' = 3x^2 - 3 = 0$ at $x = \pm 1$; only $x = 1$ is in the interval.
Critical points inside.
Candidates: $0$, $1$, $2$. Values: $f(0) = 0$, $f(1) = -2$, $f(2) = 2$.
Both ends included.
Maximum $2$ at $x = 2$; minimum $-2$ at $x = 1$. One at an end, one inside.
Match each derivative observation to the extremum conclusion.
| Local maximum | Local minimum | Neither conclusion follows without more evidence | Check critical points and endpoints | |
|---|---|---|---|---|
| $f'$ changes from positive to negative | ||||
| $f'$ changes from negative to positive | ||||
| $f'=0$ with no sign change | ||||
| Absolute extrema on $[a,b]$ |
Match each derivative fact to what it establishes about extrema.
| $c$ is a critical-point candidate | The point is also a critical-point candidate | $c$ is a local maximum | Compare all critical-point values and both endpoints | |
|---|---|---|---|---|
| $f'(c)=0$ | ||||
| $f'$ does not exist at an interior point | ||||
| $f'(c)=0$ and $f''(c)<0$ | ||||
| Find absolute extrema on $[a,b]$ |
Where is the stationary point of $f(x) = 8x^2 - 10x + 5$?
Answer:
$f(x) = |x - 3|$ on $\mathbb{R}$. At how many points is $f$ critical?
Answer:
$f(x) = 4x^3$ has $f'(0) = 0$. Is $x = 0$ a maximum, a minimum, or neither?
| Neither maximum nor minimum | Local minimum | Local maximum | More evidence is needed | |
|---|---|---|---|---|
| $f'(x)$ is positive on both sides of a critical point | ||||
| $f'(x)$ changes from negative to positive | ||||
| $f'(x)$ changes from positive to negative | ||||
| $f'(c)=0$ but there is no sign change |
$f(x) = x^3 - 9x$ on $[\,0, 2\,]$. What is its minimum value? (Take $3 = 1$ if the interval would otherwise miss the critical point.)
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f'(c) = 0$ and $f''(c) = -9$. What does the second derivative test conclude?
You can find every critical point, classify it, and apply the closed-interval method. Say in your own words why the endpoints are candidates even though Fermat's theorem says nothing about them. Next: the mean value theorem, which is where most of these facts are proved from.
8. Your turn: the extreme values of $f(x) = x^3 - 3x$ on $[0, 2]$, step 3