MA 3972-MA-Book April 11, 2018 17:21168 STEP 4. Review the Knowledge You Need to Score High
f′′(x)=
−68(x^2 −25)^2 −2(x^2 −25)(2x)(− 68 x)
(x^2 −25)^4=
68(3x^2 +25)
(x^2 −25)^3.
Setf′′>0. Since (3x^2 +25)>0,
⇒(x^2 −25)^3 > 0 ⇒x^2 − 25 >0,
x<−5orx>5. Thus, f(x) is concave
upward on (−∞,−5) and (5,∞) and
concave downward on (−5, 5).- Step 1: f(x)=x+sinx,
f′(x)= 1 +cosx,
f′′ =−sinx.
Step 2: Setf′′(x)= 0 ⇒−sinx=0or
x=0,π,2π.
Step 3: Check the intervals.
0 π 2 π[]
concave
upwardconcave
downwardf′′ + –fStep 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 /^2f′(0)=0 andf′′(0)=1
5
(andf(0)=5)⇒(0,5)isa
relative maximum. Sincef(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
extremum. Thus, (0,5) is an
absolute maximum point, and 5 is
the maximum value. Now we
check the endpoints,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 Af′< 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. - 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.