Page 90 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 4] DERIVATIVES 81

f (x)
f ð þ hÞ f ð Þ
If h > 0, then @ 0 and
h
lim f ð þ hÞ f ð Þ @ 0 ð1Þ
h
h!0þ
M
If h < 0, then f ð þ hÞ f ð Þ A 0 and
h
lim f ð þ hÞ f ð Þ A 0 x
h ð2Þ a ξ b
h!0
But by hypothesis f ðxÞ has a derivative at all points Fig. 4-9
in ða; bÞ. Then the right-hand derivative (1) must be
equal to the left-hand derivative (2). This can happen only if they are both equal to zero, in which case
f ð Þ¼ 0as required.
0
A similar argument can be used in case M ¼ 0 and m 6¼ 0.
4.20. Prove the mean value theorem.
f ðbÞ f ðaÞ
.
b a
Define FðxÞ¼ f ðxÞ f ðaÞ ðx aÞ
Then FðaÞ¼ 0 and FðbÞ¼ 0.
Also, if f ðxÞ satisfies the conditions on continuity and differentiability specified in Rolle’s theorem, then
FðxÞ satisfies them also.
Then applying Rolle’s theorem to the function FðxÞ,we obtain
f ðbÞ f ðaÞ f ðbÞ f ðaÞ
¼ 0; a < < b or ; a < < b
0 0 0
b a b a
F ð Þ¼ f ð Þ f ð Þ¼
2
4.21. Verify the mean value theorem for f ðxÞ¼ 2x 7x þ 10, a ¼ 2, b ¼ 5.
0
f ð2Þ¼ 4, f ð5Þ¼ 25, f ð Þ¼ 4 7. Then the mean value theorem states that 4 7 ¼ð25 4Þ=ð5 2Þ
or ¼ 3:5. Since 2 < < 5, the theorem is verified.
4.22. If f ðxÞ¼ 0at all points of the interval ða; bÞ, prove that f ðxÞ must be a constant in the interval.
0
Let x 1 < x 2 be any two different points in ða; bÞ. By the mean value theorem for x 1 < < x 2 ,

¼ f ð Þ¼ 0
f ðx 2 Þ f ðx 1 Þ
0
x 2 x 1
Thus, f ðx 1 Þ¼ f ðx 2 Þ¼ constant. From this it follows that if two functions have the same derivative at all
points of ða; bÞ,the functions can only differ by a constant.

4.23. If f ðxÞ > 0at all points of the interval ða; bÞ, prove that f ðxÞ is strictly increasing.
0
Let x 1 < x 2 be any two different points in ða; bÞ.By the mean value theorem for x 1 < < x 2 ,
f ðx 2 Þ f ðx 1 Þ
¼ f ð Þ > 0
0
x 2 x 1
Then f ðx 2 Þ > f ðx 1 Þ for x 2 > x 1 , and so f ðxÞ is strictly increasing.
b a 1 1 b a
4.24. (a) Prove that 2 < tan b tan a < 2 if a < b.
1 þ b 1 þ a
3 1 4 1
(b) Show that þ < tan < þ .
4 25 3 4 6
2
2
(a)Let f ðxÞ¼ tan 1 x. Since f ðxÞ¼ 1=ð1 þ x Þ and f ð Þ¼ 1=ð1 þ Þ,we have by the mean value
0
0
theorem