James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A44
appendix F Proofs of Theorems
D
We approximate the arc AB by an inscribed polygon consisting of n equal line segments
and we look at a typical segment PQ. We extend the lines OP and OQ to meet
AD in the points R and S. Then we draw RT i PQ as in Figure 2. Observe that
/RTO − /PQO , 908
B
Q T
° °
S
and so /RTS . 908. Therefore we have
| PQ | , | RT | , | RS |
¨
° °
P
R
If we add n such inequalities, we get
L n , | AD | − tan
O 1
A
where L n is the length of the inscribed polygon. Thus, by Theorem 2.3.2, we have
FIGURE 2
lim
nl ` Ln < tan
But the arc length is defined in Equation 8.1.1 as the limit of the lengths of inscribed
polygons, so
− lim
n l `
L n < tan
■
Section 4.3
Concavity Test
(a) If f 0sxd . 0 for all x in I, then the graph of f is concave upward on I.
(b) If f 0sxd , 0 for all x in I, then the graph of f is concave downward on I.
y
y=ƒ
Proof of (a) Let a be any number in I. We must show that the curve y − f sxd lies
above the tangent line at the point sa, f sadd. The equation of this tangent is
y − f sad 1 f 9sadsx 2 ad
ƒ
f(a)+f ª(a)(x-a)
So we must show that
f sxd . f sad 1 f 9sadsx 2 ad
0
FIGURE 3
a
x
x
whenever x [ I sx ± ad. (See Figure 3.)
First let us take the case where x . a. Applying the Mean Value Theorem to f on
the interval fa, xg, we get a number c, with a , c , x, such that
1 f sxd 2 f sad − f 9scdsx 2 ad
Since f 0 . 0 on I, we know from the Increasing/Decreasing Test that f 9 is increasing
on I. Thus, since a , c, we have
f 9sad , f 9scd
and so, multiplying this inequality by the positive number x 2 a, we get
2 f 9sadsx 2 ad , f 9scdsx 2 ad
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