Extreme Value Theorem, Global Versus Local Extrema, and Critical Points: Study Guide (Calculus AB 5.2)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation and words

Extremum (plural extrema) means a maximum or a minimum. Two pairs of words mean the same thing:

• global = absolute: the highest or lowest value on the whole interval or domain;
• local = relative: the highest or lowest value compared with nearby points only.

The maximum value is a y-value, f(c). It occurs at an x-value, c. Keep these apart in written answers: “the maximum value is 16, at x = 4”.

The idea: a closed journey has a highest point

A fictional hiker walks a trail for 3 hours. Her altitude changes continuously. Somewhere on the walk she was at her highest point, and somewhere at her lowest. That sounds obvious, but it needs two things: the altitude cannot jump, and the walk includes its start and its finish.

The Extreme Value Theorem (EVT) turns this into mathematics. It is the third existence theorem in the course, after the Intermediate Value Theorem (Topic 1.16) and the Mean Value Theorem (Topic 5.1).

The Extreme Value Theorem, part by part

Extreme Value Theorem. If f is continuous on the closed interval [a, b], then f has at least one maximum value and at least one minimum value on [a, b].

Part	What it says	How you check it
Condition	f is continuous on the closed interval [a, b]	Name the function type, use a given fact, or check every join and every point where a formula is undefined
Conclusion	Numbers c and d in [a, b] exist with f(d) ≤ f(x) ≤ f(c) for every x in [a, b]	Name the theorem; the extrema may be at endpoints or inside

Both parts of the condition matter.

• Open intervals break it. On (0, 1), f(x) = x has no maximum. It gets as close to 1 as you like, but never reaches 1, because x = 1 is not allowed.
• Discontinuities break it. f(x) = 1/x² on [−1, 1] is undefined at 0 and grows without bound near 0. It has no maximum value.

No guarantee is not the same as impossible. f(x) = x² on the open interval (−1, 2) still has a minimum value, 0 at x = 0. The theorem just did not promise it.

The theorem says nothing about where the extrema are, or how many times each is reached. That is the job of the critical points below and of Topic 5.5.

Global versus local extrema

Let c be in the domain of f.

• f(c) is a global maximum on an interval if f(c) ≥ f(x) for every x in the interval. (Global minimum: f(c) ≤ f(x).)
• f(c) is a local maximum if f(c) ≥ f(x) for every x in some open interval around c. (Local minimum: f(c) ≤ f(x).)

A local maximum is a “hilltop”: higher than its neighbours on both sides. A global maximum is the highest point of all. A point can be both.

Because the local definition needs an open interval around c, an endpoint is not a local extremum under this definition. It can still be a global one.

Figure 1 shows three facts worth remembering. A global extremum can sit at an endpoint. A global extremum value can be reached at more than one x. And a local maximum need not be the highest point: here the local maximum value 0 is far below the global maximum value 16.

Critical points

Critical point. A number c in the domain of f is a critical point of f if f′(c) = 0 or f′(c) does not exist.

The key link with extrema is this:

If f has a local extremum at c, then c is a critical point of f.

Why it is true. Suppose f has a local maximum at c (so f is defined on both sides of c) and f′(c) exists. For x just to the right of c, f(x) ≤ f(c), so the difference quotient (f(x) − f(c))/(x − c) is ≤ 0. For x just to the left, the top is ≤ 0 and the bottom is negative, so the quotient is ≥ 0. The derivative is the limit of both, so f′(c) is both ≤ 0 and ≥ 0. Hence f′(c) = 0. If f′(c) does not exist, c is a critical point anyway.

The converse is false. A critical point is only a candidate. Figure 2 shows the three standard cases.

How to find critical points.

1. Find f′(x) and simplify it into a single fraction or a factored form.
2. Solve f′(x) = 0 (set the top of the fraction to 0).
3. Find where f′(x) does not exist (the bottom is 0, or a corner or join in a piecewise function).
4. Keep only x-values in the domain of f. If f itself is undefined at c, then c is not a critical point.

Deciding whether each critical point is a maximum, a minimum or neither comes in Topics 5.4 and 5.7.

Worked example 1: a critical point that is not an extremum

Question. Let f(x) = x⁴ − 4x³. Find the critical points of f, and show that f has no local extremum at x = 0.

1. Differentiate. f′(x) = 4x³ − 12x² = 4x²(x − 3).
2. Where f′ = 0. 4x²(x − 3) = 0 gives x = 0 or x = 3.
3. Where f′ does not exist. f′ is a polynomial, so it exists everywhere. No extra points.
4. Critical points: x = 0 and x = 3.
5. Test x = 0 with nearby values. f(0) = 0. Write f(x) = x³(x − 4). Just left of 0 (say x = −0.1), x³ < 0 and x − 4 < 0, so f(x) > 0: f(−0.1) = 0.0041. Just right of 0 (say x = 0.1), x³ > 0 and x − 4 < 0, so f(x) < 0: f(0.1) = −0.0039. Points arbitrarily close to 0 lie both above and below f(0).

Answer. The critical points are x = 0 and x = 3. At x = 0 there is no local extremum, because f takes values larger than f(0) on the left and smaller on the right. (You will see in Topic 5.4 that x = 3 gives a local minimum, f(3) = −27.)

Why this matters. The factor x² makes f′ zero at 0 without changing sign: f′ is negative on both sides (f′(−0.5) = −3.5, f′(0.5) = −2.5). The graph flattens and keeps falling, like x³ in Figure 2 but downhill.

Worked example 2: a critical point where f′ does not exist

Question. Let g(x) = x^(2/3)(x − 10), where x^(2/3) means the square of the cube root of x. Find all critical points of g.

1. Expand to use the power rule: g(x) = x^(5/3) − 10x^(2/3).
2. Differentiate. g′(x) = (5/3)x^(2/3) − (20/3)x^(−1/3).
3. Write as one fraction. Multiply the first term by x^(1/3)/x^(1/3): g′(x) = (5x − 20)/(3x^(1/3)) = 5(x − 4)/(3x^(1/3)), for x ≠ 0.
4. Where g′ = 0. The top is 0 when x = 4.
5. Where g′ does not exist. The bottom is 0 when x = 0. Check the domain: g(0) = 0 × (−10) = 0, so g is defined at 0.
6. Critical points: x = 0 and x = 4.

Interpretation at x = 0. For x close to 0 (on either side), x^(2/3) > 0 and x − 10 < 0, so g(x) < 0 = g(0). For example, g(−1) = −11 and g(1) = −9. So g has a local maximum value of 0 at x = 0, at a cusp where the derivative does not exist. If you had only solved g′(x) = 0, you would have missed it.

At x = 4. g(4) = 4^(2/3)(−6) = −12∛2 ≈ −15.12. Its type is decided in Topic 5.4.

Common slip. Cancelling or ignoring x^(1/3) in the denominator loses the critical point at 0. Always look at the bottom of f′ as well as the top.

Worked example 3: does the Extreme Value Theorem apply?

Question. For each function, say whether the Extreme Value Theorem guarantees both a maximum value and a minimum value on the given interval. Explain.

(a) f(x) = 1/(x − 2) on [3, 5] (b) f(x) = 1/(x − 2) on [0, 3] (c) f(x) = x² on (−1, 2)

(a) Yes. The only discontinuity of f is at x = 2, which is outside [3, 5]. So f is continuous on the closed interval [3, 5], and the theorem applies. (The values are f(3) = 1 and f(5) = 1/3; f decreases, so these are the maximum and minimum.)

(b) No guarantee. x = 2 is inside [0, 3], and f is undefined there. The condition fails. In fact f grows without bound as x → 2 from the right, so there is no maximum.

(c) No guarantee. The interval is open. f has a minimum value, 0 at x = 0, but no maximum: f(x) gets close to 4 as x → 2, yet x = 2 is not in the interval. On the closed interval [−1, 2], the maximum value 4 would be reached at x = 2.

What to write. A full answer names the condition (continuous on a closed interval), says whether it holds with a reason, and only then names the theorem.

Common misconceptions

• “A critical point is always a maximum or minimum.” x³ and ∛x at 0 are counterexamples.
• Only solving f′(x) = 0. Points where f′ does not exist are critical points too, if f is defined there.
• Including points outside the domain. If f(c) is undefined, c is not a critical point.
• “The global maximum is at a critical point.” It can be at an endpoint instead (Figure 1).
• Mixing up value and location. “The maximum is x = 4” is wrong; “the maximum value is 16, at x = 4” is right.
• “The theorem says the extremum happens once.” A maximum value can be reached at several x-values.
• Applying the theorem on an open interval or across a break. Check closed and continuous every time.
• “No guarantee, so no maximum.” The theorem can be silent while an extremum still exists.

Where this leads

Critical points are the starting point for the rest of Unit 5. In Topic 5.3, Determining Intervals on Which a Function Is Increasing or Decreasing, you will use the sign of f′ between critical points. Topic 5.4 classifies critical points with the First Derivative Test, and Topic 5.5 combines the Extreme Value Theorem with critical points to find global extrema. For the earlier existence theorem, see Topic 5.1, Using the Mean Value Theorem. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.