JALILAH RUBAI
Kolej Matrikulasi Johor
Relative maximum/minimum points, increasing/decreasing, inflection points, concave up/down
y
> 0 – increase
= 0 s >0 < 0 – decrease
<0 < 0 r < 0 – concave down
a
> 0 > 0 > 0
=0
= ( )
q <0
=0
c d =0
>0
b < 0 ex
> 0 > 0 – concave up
p
= 0
= 0 (a, s) – maximum point; (c, p) – minimum point; observe behaviour of on the left and right of the min/max points.
= 0 (b, q), (d, q) and (e, r) – inflection points; observe the behaviour of on the left and right of inflection points.
Maximum/minimum Point
1. Find .
2. Find x (stationary values) when = 0. Increasing/decreasing
3. Test > 0 – the function is increasing
i) first derivative test (there is a change of signs)
< 0 – the function is decreasing
c ccc cd
′( ) + ′( ) + ′( ) ′( ) + + ′( ) + +
( ) is increasing on the interval (−, )
maximum minimum not a relative extrema and , .
( ) is decreasing on the interval ( , ).
or ii) second derivative test
> 0 – minimum occurs
< 0 – maximum occurs
= 0 – nor minimum neither maximum occurs
4. Conclude: ( , ) is the maximim/minimum point or not.
Inflection Point Concavity
1. Find . > 0 – the function is concave up
2. Find x when = 0. < 0 – the function is concave down
3. Test (there is a change of signs) ab
a aaa "( ) + +
"( ) + "( ) + "( ) "( ) + + ( ) is concave up on the interval (−, )
and , .
( ) is concave down on the interval ( , ).
inflection point not an inflection point
4. Conclude: ( , ) is the inflection point or not.
Example 1: Given = 2 − 4 + 3. Find 1
a) the maximum and/or minimum point(s).
b) the interval on which is increasing or decreasing. ′( ) +
c) the inflection point(s) and determine the concavity. "( ) + +
Hence, sketch the graph of .
y
Solution: c) " = 4
1 x
a) = 4 − 4 " ≠ 0 x 1
4 − 4 = 0 Therefore there is no
= 1 inflection point.
1 " > 0 x
Therefore is concave up
′( ) + on the interval −, .
1 = 2(1) −4 1 + 3 = 1
(1, 1) is the minimum point.
b) is increasing on 1,
and decreasing on −, 1 .
Example 2: Given = − 6 + 9 + 3. Find 12 3
a) the maximum and/or minimum point(s).
b) the interval on which is increasing or decreasing. ′( ) + +
c) the inflection point(s) and determine the concavity. "( ) + +
Hence, sketch the graph of .
y
Solution: b) is increasing on −, 1 and 7
3, , and decreasing on 1, 3 . 5
a) = 3 − 12 + 9 3
= 3 − 4 + 3 c) " = 6 − 12
= 3 − 3 − 1 6 − 12 = 0 123 x
= 2
3 − 3 − 1 = 0
= 1, 3 2
13 "( ) +
′( ) + +
2 = 5
(2, 5) is the inflection point.
1 = 7, 3 = 3
(1, 7) is the maximum point and is concave up on 2, and
(3, 3) is the minimum point. concave down on −, 2 .
Example 3: Given = ( − 2) . Find ′( ) 2
a) the maximum and/or minimum point(s). "( )
b) the interval on which is increasing or decreasing. ++
c) the inflection point(s) and determine the concavity. +
Hence, sketch the graph of .
Solution: c) " = 12( − 2) × 1 y
= 12( − 2)
a) = 3( − 2) × 1 x
= 3( − 2) 12( − 2)= 0
= 2 2
3( − 2) = 0
= 2 2
2 "( ) +
′( ) + +
2 = 0
(2, 0) is the inflection point.
There is no relative extrema.
is concave up on 2, and
b) is increasing on the concave down on −, 2 .
interval −, .
Example 4: Given = − + 3 − 4. Find 01 2
a) the maximum and/or minimum point(s).
b) the interval on which is increasing or decreasing. ′( ) + +
c) the inflection point(s) and determine the concavity. "( ) + +
Hence, sketch the graph of .
y
Solution: b) is decreasing on −, 0 and x
2, , and increasing on 0, 2 . 1 2
a) = −3 + 6 2
= −3 − 2 c) " = −6 + 6 4
−6 + 6 = 0
−3 − 2 = 0 = 1
= 0, 2
1
02
"( ) +
′( ) +
1 = −2
0 = −4, 2 = 0 (1, 2) is the inflection point.
(0, 4) is the maximum point and
(2, 0) is the minimum point. is concave down on 1, and
concave down on −, 1 .
With the knowledge of inequallities, limits and differentiation,
you can sketch graph of any function. The functions that are
being discussed here are limited to polynomials up to degree
three.
Thank You