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UAE-Math-Grade-12-Vol-1-SE-718383-ch4
4-43 QC: OSO/OVY T1: OSO October 25, 2018 SECTION 4.5 • • Concavity and the Second Derivative Test 253
- 0 + Solution Here, we have
4x
0 f (x) = 4x − 12x = 4x(x − 3)
′
3
2
- 0 +
( x - √ 3) √ √
√ 3 = 4x(x − 3)(x + 3).
- 0 +
′
( x + √ 3) We have drawn number lines for the factors of f (x) in the margin. From this, we
- √ 3 can see that
- 0 + 0 - 0 +
f�(x) √ √
- √ 3 0 √ 3 { > 0, on (− 3, 0) ∪ ( 3, ∞) f increasing.
′
f (x)
√
- 0 + < 0, on (−∞, − 3) ∪ (0, √ 3). f decreasing.
12(x - 1)
1 ′′ 2
Next, we have f (x) = 12x − 12 = 12(x − 1)(x + 1).
- 0 +
(x + 1)
-1 We have drawn number lines for the two factors in the margin. From this, we can
+ 0 - 0 - see that
f�(x)
-1 1 { > 0, on (−∞, −1) ∪ (1, ∞) Concave up.
′′
f (x)
- 0 + 0 - 0 + < 0, on (−1, 1). Concave down.
f'(x)
-√3 0 √3 For convenience, we have indicated the concavity below the bottom number line,
+ 0 - 0 + with small concave up and concave down segments. Finally, observe that since the
f''(x)
-1 1 graph changes concavity at x =−1 and x = 1, there are inflection points located at
(−1, −4) and (1, −4). Using all of this information, we are able to draw the graph
y shown in Figure 4.57. For your convenience, we have reproduced the number lines
′
′′
for f (x) and f (x) together above the figure.
10
′′
As we see in example 5.3, having f (x) = 0 does not imply the existence of an
inflection point.
5
x EXAMPLE 5.3 A Graph with No Inflection Points
-3 3
4
Determine the concavity of f(x) = x and locate any inflection points.
-5
′
3
Solution There’s nothing tricky about this function. We have f (x) = 4x and
′
′′
′
2
-10 f (x) = 12x . Since f (x) > 0 for x > 0 and f (x) < 0 for x < 0, we know that f
′′
is increasing for x > 0 and decreasing for x < 0. Further, f (x) > 0 for all x ≠
′′
FIGURE 4.57 0, while f (0) = 0. So, the graph is concave up for x ≠ 0. Further, even though
′′
y = x − 6x + 1 f (0) = 0, there is no inflection point at x = 0. We show a graph of the function
2
4
in Figure 4.58.
y
We now explore a connection between second derivatives and extrema. Suppose
′
that f (c) = 0 and that the graph of f is concave down in some open interval containing
c. Then, near x = c, the graph looks like that in Figure 4.59a and hence, f(c) is a local
4
′
maximum. Likewise, if f (c) = 0 and the graph of f is concave up in some open interval
containing c, then near x = c, the graph looks like that in Figure 4.59b and hence, f(c)
is a local minimum.
2
y y
x f (c) = 0
-2 -1 1 2 f�(c) > 0
FIGURE 4.58
y = x 4
Copyright © McGraw-Hill Education
f (c) < 0 f�(c) = 0
x x
c c
FIGURE 4.59a FIGURE 4.59b
Local maximum Local minimum
272 | Lesson 4-5 | Concavity and the Second Derivative Test