1. Relative Extrema
First Derivative Test - FDT
Second Derivative Test - SDT

2. Relative Extrema
de
c
x
1
relative
maximum
re
as
ing
x
g
sin
a
cre
in
2
relative
minimum
de
cre
as
x
3
relative
maximum
ing
x
4
relative
minimum

3. Critical Points
Critical points are points at which:
•Derivative equals zero (also called
stationary point).
•Derivative doesn’t exist.

4. First Derivative Test
Let f be a differentiable function with f '(c) = 0, then:
•If f '(x) changes from positive to negative, then f has a
relative maximum at c.
•If f '(x) changes from negative to positive, then f has a
relative minimum at c.
•If f '(x) has the same sign from left to right, then f
does not have a relative extremum at c.

5. Practice Time!!!
Use First Derivative Test to find critical
points and state whether they are
5
2
minimums or maximums. f ( x ) = 3x 3 −15x 3
2
−1
−1
2
1
−
5 3
2 3
3
3
f ' ( x ) = 3× x −15× x
f ' ( x ) = 5x −10x = 5x 3 ( x − 2 )
3
3
1
−
5 ( x − 2)
5x 3 ( x − 2 ) = 0
Critical points
=0
1
x3
+
0
__
•
0
relative
maximum
•
2
relative
minimum
+
2(stationary)

6. Second Derivative Test
Suppose that c is a critical point at which f’(c) = 0,
that f(x) exists in a neighborhood of c, and that f(c)
exists. Then:
• f has a relative maximum value at c if f”(c) < 0.
•f has a relative minimum value at c if f”(c) > 0.
•If f(c) = 0, the test is not conclusive.
Note: Second derivative test is still
used to calculate max and min

7. Practice Time again !!!
Use second derivative test to find extrema
of f (x) = 3x 5 − 5x 3
f '(x) = 15x − 15x
4
2
15x − 15x = 0
4
2
critical points = 0, -1, 1
f "(x) = 60x 3 − 30x
inconclusive
f "(0) = 0
f "(− 1) = − 30 < 0 f has a maximum at x = -1
f has a minimum at x = 1
f "(1) = 30 > 0
15x 2 ( x 2 − 1) = 0

8. x2
Find extrema of f (x) =
x 4 +16
A.Using first derivative test
B.Using second derivative test