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UAE_Math_Grade_12_Vol_1_SE_718383_ch3
It turns out that our earlier observation regarding the location of extrema is correct.
That is, local extrema occur only at points where the derivative is zero or undefined.
We state this formally in Theorem 3.2.

THEOREM 3.2 (Fermat’s Theorem)
Suppose that f(c) is a local extremum (local maximum or local minimum). Then c
must be a critical number of f.

PROOF
Suppose that f is differentiable at x = c. (If not, c is a critical number of f and we are
′
′
′
done.) Suppose further that f (c) ≠ 0. Then, either f (c) > 0 or f (c) < 0.
′
If f (c) > 0, we have by the definition of derivative that
f(c + h) − f(c)
′
f (c) = lim > 0.
h→0 h
So, for all h sufficiently small,
f(c + h) − f(c) > 0. (3.1)
TODAY IN h
MATHEMATICS For h > 0, (3.1) says that

Andrew Wiles (1953– ) f(c + h) − f(c) > 0
A British mathematician who
in 1995 published a proof of and so,
Fermat’s Last Theorem, the most
famous unsolved problem of the f(c + h) > f(c).
20th century. Fermat’s Last Thus, f(c) is not a local maximum.
Theorem states that there is no Similarly, for h < 0, (3.1) says that
integer solution x, y and z of the
n
equation x + y = z for f(c + h) − f(c) < 0
n
n
integers n > 2. Wiles had
wanted to prove the theorem and so,
since reading about it as a
10-year-old. After more than ten f(c + h) < f(c).
years as a successful research Thus, f(c) is not a local minimum, either.
mathematician, Wiles isolated Since we had assumed that f(c) was a local extremum, this is a contradiction. This
himself from colleagues for rules out the possibility that f (c) > 0.
′
seven years as he developed the ′
mathematics needed for his We leave it as an exercise to show that if f (c) < 0, we obtain the same contradiction.
′
proof. “I realised that talking to The only remaining possibility is to have f (c) = 0 and this proves the theorem.
people casually about Fermat
was impossible because it We can use Fermat’s Theorem and calculator- or computer-generated graphs to
generated too much interest. find local extrema, as in examples 3.6 and 3.7.
You cannot focus yourself for
years unless you have this kind
of undivided concentration
which too many spectators EXAMPLE 3.6 Finding Local Extrema of a Polynomial
Copyright © McGraw-Hill Education intense work on this one step, as Solution Here, f (x) = 6x − 6x − 12 = 6(x − x − 2)
would destroy.” The last step of
3
2
Find the critical numbers and local extrema of f(x) = 2x − 3x − 12x + 5.
his proof came, after a year of
“this incredible revelation” that
′
was “so indescribably beautiful,
2
2
it was so simple and elegant.”
= 6(x − 2)(x + 1).
253