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Relative Maxima and Minima
sections 4.
Definition. By a critical point of a function f we mean a point x 0 in the domain at which
either the derivative is zero or it does not exists. So , geometrically , one of the following cases
happen at a critical point:
(i) the tangent line is horizontal
(ii) the tangent line is vertical (this corresponds to the case where the derivative is one of 1)
(iii) the tangent line does not exist ; there is a cusp on the graph at x 0
Example. Find the critical points of the function f (x) =
x^2 x^3 1
Solution.
f
′ (x) =
(x
2 )
′ (x
3 1) (x
2 )(x
3 1)
′
(x^3 1)^2
x(x
3
(x^3 1)^2
Now we need to look for the points at which the derivative is zero or the derivative does not
exist.
f
′ (x) = 0 ) x = 0 ,
3
p 2
there is no point in the domain where f is not differentiable (note that the point x = 1 is not
in the domain at all , so it is not considered as a critical point). So , the only critical points
are x = 0 ,
3
p
Definition. A function f is said to have relative minimum (local minimum) at x 0 if there
exists some open interval I containing x 0 such that
f (x 0 ) f (x) for all x 2 I
Definition. A function f is said to have relative maximum (local maximum) at x 0 if
there exists some open interval I containing x 0 such that
f (x) f (x 0 ) for all x 2 I
Question. The following theorem tells us where to search for relative extrema ; in fact it says
that we need to llok for the critical points if we really want to search for the relative extrema.
Theorem. If a function f has a relative extrema over an interval I at a point x 0 2 I and if
x 0 is not an endpoint of I , then x 0 is a critical point of f , and therefore either f (x 0 ) = 0 or
f
′ (x 0 ) does not exist.
First-Derivative Test for Relative Extrema. Suppose that c is a critical point of a con-
tinuous function f (f
′ (c) may or may not exist).
(i) If on both sides of c we have this situation:
c
f
′
f
then f has a relative minimum at c.
(ii) If on both sides of c we have this situation:
c
f
′
f
then f has a relative maximum at c.
(iii) If on both sides of c the derivative f
′ does not change sign:
c
f
′
f
c
f
′
f
If f
′′ (c) > 0 , then c is a point of relative minimum for f.
If f
′′ (c) < 0 , then c is a point of relative maximum for f.
If f
′′ (c) = 0 , then no conclusion can be made about c.
Note. The rst-derivative test is used for both types of critical points no matter whether
f
′ (c) = 0 or f
′ (c) does not exist. But , the second-derivative test is used only for the critical
points satisfying f
′ (c) = 0. For example, in the previous example , the second-derivative test
cannot be applied for the critical point x = 0. But , it can be used to decide about x =
2
f
′ (x) =
1
3
(5x 2)x
1 (^3) )
f
′′ (x) =
1
3
(5x 2)
′ (x
1 (^3) ) +
1
3
(5x 2)(x
1 (^3) )′
=
1
3
(5)(x
1 (^3) ) +
1
3
(5x 2)(
1
3
x
4 (^3) ) =
5
3
3
p x
5 x 2
9 x
3
p x
=
15 x (5x 2)
9 x
3
p
x
=
10 x + 2
9 x
3
p
x
) f
′′ (
2
5
) > 0 )
2
5
gives relative minimum
Example (section 4.3 exercise 19). Find the the critical points of the
function f (x) = (x + 2)
3 (x 4)
3 , and determine which critical points give
relative maxima and which ones give relative minima.
Solution.
f
′ (x) = f(x + 2)
3 g
′ f(x 4)
3 g + f(x + 2)
3 gf(x 4)
3 g
′
= f3(x + 2)
2 gf(x 4)
3 g + f(x + 2)
3 gf3(x 4)
2 g
= 3(x + 2)
2 (x 4)
2 f(x 4) + (x + 2)g
= 3(x + 2)
2 (x 4)
2 f 2 x 2 g
= 6(x + 2)
2 (x 4)
2 (x 1)
f
′ (x) = 0 ) x = 2 , 1 , 4 (the only critical points)
(there are no critical points at which f
′ does not exist).
The only term whose sign must be determined is the term (x 1) because
the signs of the other terms don’t change.
1 4
x − 1
f − − + +
′
f
− 2
relative
min
x
f
′
relative minimum
f
