MA 3972-MA-Book April 11, 2018 17:21
Graphs of Functions and Derivatives 171
Step 9: Sketch the graph.
(See Figure 8.8-4.)
x
y
f
x = 4
y = 1
4
1
Figure 8.8-4
- (a)
x 1 x 4
++
rel. max. rel. min.
incr. decr. incr.
f′ –
f
The functionfhas the largest value
(of the four choices) atx=x 1.
(See Figure 8.8-5.)
0 x 1 x
f
y
x 2 x 3 x 4
Figure 8.8-5
(b) Andfhas the smallest value atx=x 4.
(c)
x 3 +
decr. incr.
f′
f′′
f concave
upward
pt. of
inflection
concave
downward
A change of concavity occurs atx=x 3 ,
and f′(x 3 ) exists, which implies there
is a tangent to fatx=x 3. Thus, at
x=x 3 , fhas a point of inflection.
(d) The functionf′′represents the slope
of the tangent tof′. The slope of the
tangent to f′is the largest atx=x 4.
- (a) Since f′(x) represents the slope of the
tangent,f′(x)=0atx=0, andx=5.
(b) Atx=2,f has a point of inflection,
which implies that iff′′(x) exists then
f′′(x)=0. Since f′(x) is differentiable
for all numbers in the domain,f′′(x)
exists, andf′′(x)=0atx=2.
(c) Since the functionf is concave
downward on (2,∞),f′′<0on
(2,∞), which implies f′is decreasing
on (2,∞).
- (a) The function fis increasing on the
intervals (−2, 1) and (3, 5) and
decreasing on (1, 3).
(b) The absolute maximum occurs at
x=1, since it is a relative maximum,
f(1)> f(−2) andf(5)< f(−2).
Similarly, the absolute minimum
occurs atx=3, since it is a relative
minimum, andf(3)< f(5)< f(−2).
(c) No point of inflection. (Note that at
x=3,f has a cusp.)
Note that some textbooks define a
point of inflectionas a point where the
concavity changes and do not require
the existence of a tangent. In that case,
atx=3,f has a point of inflection.
(d) Concave upward on (3, 5) and
concave downward on (−2, 3).
(e) A possible graph is shown in
Figure 8.8-6.
