Absolute (global) extrema. For f defined on a domain D, f(x0) is the absolute maximum of f on D if f(x0)≥f(x) for every x∈D; the absolute minimum is defined symmetrically with ≤.
Extreme Value Theorem. If f is continuous on a closed interval [a,b], then f attains both an absolute maximum and an absolute minimum somewhere on [a,b] — and the extremum can only occur either at an interior critical number or at one of the two endpoints.
Procedure for absolute extrema on [a,b] (Exercise 7.6 Q1's method):
Find every critical number of f in the open interval (a,b).
Evaluate f at each critical number and at both endpoints a,b.
The largest of these values is the absolute maximum; the smallest is the absolute minimum.
Relative (local) extrema.f has a relative (local) maximum at x0 if f(x0) is the largest value of f on some open interval around x0 (relative minimum: smallest, on some open interval). A function may have several local extrema, and a local extremum need not be the absolute one.
Fermat's Theorem. If f has a relative extremum at x=c, then c must be a critical number of f (so the search for local extrema always starts by solving f′(x)=0 together with any points where f′ fails to exist) — though not every critical number is automatically an extremum (e.g. y=x3 at x=0).
First Derivative Test. At a critical point c where f is continuous, examine the sign of f′(x) moving left to right across c:
negative → positive: local minimum at c;
positive → negative: local maximum at c;
no sign change (same sign on both sides): c is neither a local max nor a local min.
Second Derivative Test (an alternative, often quicker, at a stationary point). If f′(c)=0 and f′′(c) exists:
f′′(c)<0⇒ local maximum at c;
f′′(c)>0⇒ local minimum at c;
f′′(c)=0⇒ the test is inconclusive — fall back to the first derivative test.
Optimization (applied maxima/minima). A real-world "find the maximum/minimum ___" word problem follows the same five steps every time: (1) draw a figure and label the relevant quantities; (2) write an expression for the quantity to be extremised; (3) use the problem's constraint to reduce that expression to a single variable; (4) determine the valid interval of that variable from the physical setup; (5) apply absolute extrema, the first-, or the second-derivative test to obtain the answer — then translate the critical value(s) back into the original quantities the question asked for.
Tip
Whenever a constraint relates two variables (e.g. xy=k or x+y=S), eliminate one of them before differentiating — optimizing a two-variable expression directly is a much harder (Lagrange-multiplier) problem that this chapter does not need.
Find critical numbers in each open interval, evaluate f there and at both endpoints, compare.
✓Final answer
Max −1 at x=1, min −10 at x=2.
Max 16 at x=2, min −1 at x=1.
Max 9 at x=−1, min −89 at x=81.
Max 233 at x=6π, min 0 at x=2π.
In each part, find the critical numbers inside the given closed interval, evaluate f at those and at both endpoints, then read off the largest/smallest.
Step 1 (i). f(x)=x2−12x+10 on [1,2].
f′(x)=2x−12=0⇒x=6∈/(1,2) — no interior critical number. Evaluate at endpoints: f(1)=1−12+10=−1; f(2)=4−24+10=−10.
Max =−1 at x=1; min =−10 at x=2.
Step 2 (ii). f(x)=3x4−4x3 on [−1,2].
f′(x)=12x3−12x2=12x2(x−1)=0⇒x=0,1 (both in (−1,2)).
f(−1)=3+4=7; f(0)=0; f(1)=3−4=−1; f(2)=48−32=16.
Max =16 at x=2; min =−1 at x=1.
Step 3 (iii). f(x)=6x4/3−3x1/3 on [−1,1].
f′(x)=8x1/3−x−2/3=x−2/3(8x−1). Critical numbers: x=0 (f′ undefined) and x=1/8 (8x−1=0), both in (−1,1).
(i) Absolute max −1 (at x=1), absolute min −10 (at x=2). (ii) Max 16 (at x=2), min −1 (at x=1). (iii) Max 9 (at x=−1), min −89 (at x=81). (iv) Max 233 (at x=6π), min 0 (at x=2π).
Extreme Value Theorem procedure: critical numbers + endpoints, compare
Missing a critical number where f′ is undefined rather than zero (part iii)
Discarding the physically-invalid root sinx=−1 but forgetting to also check both endpoints in part (iv)