Take a Diagnostic Exam 29
[
[
[
[
a
adb
0 e b
f' +
f"
f'
0– +0
f incr.
f
decr. incr.
decr. incr.
Concave downward Concave upward
aex
y
0 d b
A possible graph of f′
Based on the graph of f:
Figure DS-2
- The graph indicates that (1)f(10)=0,
(2)f′(10)<0, sincef is decreasing; and
(3) f′′(10)>0, sincef is concave upward.
Thus,f′(10)< f(10)< f′′(10), choice (C).
- See Figure DS-3.
The graph off is concave upward for
x<x 2.
x 2
f′ incr.
f′′
f
decr.
+–
Concave
upward
Concave
downward
Figure DS-3
- See Figure DS-4.
Entery^1 =sin(x^2 ). Using the [Inflection]
function of your calculator, you obtain four
points of inflection on [0, π]. The points of
inflection occur atx= 0 .81, 1.81, 2.52, and
3.07. Sincey 1 =sin (x^2 ) is an even function,
there is a total of eight points of inflection on
[−π,π]. An alternate solution is to enter
y^2 =
d^2
dx^2
(y 1 (x),x, 2). The graph ofy 2 crosses
thex-axis eight times, thus eight zeros on
[−π,π].
Figure DS-4
- Sinceg(x)=
∫ x
a
f(t)dt, g′(x)=f(x).
See Figure DS-5.
The only graph that satisfies the behavior ofg
is choice (A).
[ [
ab^0
rel. max.
+– 0
incr. decr.
g′(x)=f(x)
g(x)
Figure DS-5
- See Figure DS-6.
A change of concavity occurs atx=0 forq.
Thus,qhas a point of inflection atx=0.
None of the other functions has a point of
inflection.
[ [
a 0 b
incr decr
+–
Concave
upward
Concave
downward
q′
q′′
q
Figure DS-6
