We
will look at a function from earlier and see how it is evaluated
for Concavity.
The
function is: ¦
(x)= 7x^{4 } 24x^{3 }+ 10x^{2} + 24x
– 17
The
1st Derivative:¦’(x)= 28x^{3 }– 72x^{2 }+ 20x + 24
The
2nd Derivative ¦”=
84x^{2} – 144x + 20 «
4(21 x^{2 }– 36x + 5)
Where
it will be zero for Concave UP È
or Concave DOWN Ç.
4(21
x^{2 }– 36x + 5) = 0 à
Quadratic
formula à
x= (36 +or (36^{2 }– 4(21)(5))^{½})/ 42
x=0.1524453
and x=1.56184 à
now set these as intervals.
This
is a table of the intervals
(¥,
0.1524),

(0.152,
1.56184)

(1.56184,
¥)

(¦’’(3)=Pos)

(¦’’(1)=Neg)

(¦’’(5)=Pos)

Concave
UP È

Concave
DOWN Ç,

Concave
UP È




Another example of
Determining Concavity:
¦
(x)= 6/(x² + 3)«
6(x² + 3)^{1
}
^{
}
¦’(x)=
(6)( x² + 3)^{2} (2x) «
12x/(x² + 3)²^{
}
¦”(x)=
[(x² + 3)²(12)(12x)(2)( x² + 3)(2x)]/( x² + 3)³
= [36(x² +1)]/(x² + 3)³
Because
¦”(x)
= 0 when x = ±
1 and ¦”
is defined on the entire real
line, you should test ¦”
on the intervals (¥,
1), (1,1), and (1,¥).
The results are shown in the table:
(¥,
1),

(1,
1)

(1,
¥)

¦’’(2) > 0

¦’’(0) < 0

(¦’’(2) > 0

Concave
UP È

Concave
DOWN Ç,

Concave
UP È




Points
of Inflection
The
Concavity of ¦
changes at a point of inflection.
THEOREM
3.8  Points of Inflection
If (c, ¦(c))
is a point of inflection on the graph of ¦,
then either ¦”(c)=
0 pr ¦
is not differentiable at x = c.
þ
¦
(x)= 3x^{4} –4x³ on interval [1,2]
þ
For Critical Numbers get 1^{st}
derivative and set to zero
þ
¦’(x)
= 12x³12x²
þ
0= 12x(x1)
therefore: x=0,x=1 (Critical Numbers)
þ
Now consider all point,
the Critical and the End points.
þ
¦(1)=
3x^{4} –4x³ = 7 Þ
(1,7)
þ
¦(0)=
3x^{4} –4x³ =
0 Þ
(0,0)
þ
¦(1)=
3x^{4} –4x³ =
1 Þ
(1, 1)Þ
Minimum
þ
¦(2)=
3x^{4} –4x³ =
16 Þ
(2,16)Þ
Maximum
þ
Now we look at derivative
of ¦’(x)
= 12x³12x² which is the second derivative of ¦(x)=
3x^{4} –4x³
þ
¦”
(x)= 12x² – 24 «
12x(x –2)
þ
Set ¦”(x)
= 0, you can determine that
the possible points of inflection occur at x=0 and x=2.
By testing the intervals determined by these x values, you
can conclude that they both yield points of inflection.
þ
Here is the Table:
þ
Interval


¥
< x < 0

0
< x < 2

2
< x < ¥

Test value

X
=  1

X
= 1

X
= 3

Sign of ¦”
(x)

¦”
(1) > 0

¦”
(1) < 0

¦”
(3) > 0

Conclusion

Concave
UP È

Concave
DOWN Ç,

Concave
UP È

Using
the 2^{nd} Derivative Test.
In addition to
testing for concavity, the 2^{nd} Derivative Test
can be used to perform a simple test for relative maxima &
minima.
THEOREM
3.9 – Second Derivative Test
Let
¦
be a function such that ¦’(c)=
0 and the 2^{nd} derivative of ¦
exists on an open interval containing c.
1.
If ¦”(c)>
0, then ¦”(c)
is a relative minimum.
2.
If ¦”(c)<
0, then ¦”(c)
is a relative maximum.
3.
If ¦”(c)=
0, the test fails. In
such cases, you use the 1^{st} Derivative Test.
Another example
using the 2^{nd} Derivative Test.
Find the relative
extrema for ¦
(x) = 3x^{5} + 5x³
Solution: Begin by
finding the critical numbers of ¦.
¦′(c)=
15x^{4} + 15x² «
15x²(1  x²)= 0
Critical numbers:
x=1,0,1
Using the 2^{nd}
Der: ¦”(c)=
60x³ + 30x «
= 30(2x³ + x)
Point

(1,
2)

(1,2)

(0,0)

Sign of ¦”
(x)

¦”
(1) > 0

¦”
(1) < 0

¦”
(0) = 0

Conclusion

Relative
minimum

Relative
maximum

Test
fails

Because the 2^{nd}
Derivative Test fails at (0,0), you can use the 1^{st}
Derivative Test and observe that ¦
é
to the left and the right of x=0.
SO, 90,0) is neither a relative min nor a relative max.
