Graphs of Functions and Derivatives 141
Step 3:Check intervals.
0 π 2 π
[]
concave
upwarddownward
concave
f ′′ +–
f
Step 4:Check for tangent line: Atx=π,
f′(x)= 1 +(−1)⇒0 there is a
tangent line atx=π.
Step 5:Thus, (π,π) is a point of
inflection.
- Step 1: Rewrite f(x)as
f(x)=(25−x^2 )^1 /^2.
Step 2: f′(x)=
1
2
(25−x^2 )−^1 /^2 (− 2 x)
=
−x
(25−x^2 )^1 /^2
Step 3:Find critical numbers. f′(x)=0;
atx=0; andf′(x) is undefined at
x=±5.
Step 4:
f′′(x)
=
(−1)
√
(25−x^2 )−
(− 2 x)(−x)
2
√
(25−x^2 )
(25−x^2 )
=
− 1
(25−x^2 )^1 /^2
−
x^2
(25−x^2 )^3 /^2
f′(0)=0 and f′′(0)=
1
5
(and f(0)=5)⇒(0,5)isa
relative maximum. Since f(x)is
continuous on [−5, 5],f(x) has
both a maximum and a minimum
value on [−5, 5] by the Extreme
Value Theorem. And since the
point (0,5) is the only relative
extremum, it is an absolute
extramum. Thus, (0,5) is an
absolute maximum point and 5 is
the maximum value. Now we
check the end points, f(−5)= 0
and f(5)=0. Therefore, (−5, 0)
and (5, 0) are the lowest points for
f on [−5, 5]. Thus, 0 is the
absolute minimum value.
- (a) Point A f′< 0 ⇒decreasing and
f′′> 0 ⇒concave upward.
(b) Point Ef′< 0 ⇒decreasing and
f′′< 0 ⇒concave downward.
(c) Points B and Df′= 0 ⇒horizontal
tangent.
(d) Point Cf′′does not exist⇒vertical
tangent.
7.A change in concavity⇒a point of
inflection. Atx=a, there is a change of
concavity;f′′goes from positive to
negative⇒concavity changes from
upward to downward. Atx=c, there is a
change of concavity;f′′goes from
negative to positive⇒concavity changes
from downward to upward. Therefore,f
has two points of inflection, one atx=a
and the other atx=c.
8.Setf′′(x)=0. Thus,x^2 (x+3)(x−5)=
0 ⇒x=0,x=−3, orx=5. (See
Figure 7.9-1.)
Thus,f has a change of concavity at
x=−3 and atx=5.
Figure 7.9-1
