Answers & Hints for Selected Exercises 671Case 2: x1 < xo. Revise the above argument to cover this case.- (a) Suppose f is differentiable on (xo - 8,xo), continuous from the left at
Xo and lim f'(x) exists. Let x E (xo - 8,xo). Applying the MVT to f on
x-+xQ
[x ) X 0 ] ) :Jc x E (x ' xo)^3 f'(c x ) = f(x)-f(xo). x-xo Thus ) -f' (x 0 ) = lim - f(x)-f(xo) x-xo =
X-+X 0
lim f'(cx) = lim f'(cx) = lim f'(x), which exists by hypothesis.
x -+xQ Cx-+xQ x--+xQ
.'. f'(xo) exists and equals lim f'(x).
1 2 1. 1. x^2 sin~-f(O)- For x =f=. 0, f (x) = --x cos ::2 x + 2x sm ::2x • Also, f'(O) = x-+O hm x _ 0 =
lim xsin ~ = 0 by Ex. 4.2.19. Thus, f is differentiable everywhere.
x--+O x
Let Xn = 1/y"I;ITTi. Then Xn---+ 0 but f'(xn) = -2y"I;ITTi---+ oo. Therefore,
Ve> 0, f is differentiable on [-c,c] but f' is unbounded there. - See solution to Exercise 6.1.17 and apply Thm. 6.4.4.
EXERCISE SET 6.5- T 4 (x) = 5 + 2-(x - 1) - (x - 1)^2 + (x - 1)^4 = 5 + 2x - 2 - (x^2 - 2x + 1) +
( x^4 - 4x^3 + 6x^2 - 4x + 1) = 3 + 5x^2 - 4x^3 + x^4. - T2n+i ( X ) -- X - 3T x3 + ST xs - · · · + ( -l)n (2n+l)!' x2n+1.
T2n ( X ) -- X - 3T x3 +ST xs - · · · + -( l)n-1 (2n-l)! x2n-1 -- rri -'-2n-1 ( X ) ·- T2n(x) = 1 - ~~ + ~~ - ~f + .. · + (-lr (~:~,; T2n+i (x) = T2n(x).
- T6(x) = 1 + ~(x - 1) - 2 l 21 (x - 1)^2 + 2 l 31 (x - 1)^3 - 2 l_~ 1 (x - 1)^4
+~~s5~(x -1)5 - 32t~·,9(x -1)6.
Let x =f=. 1. R 6 (x) = 1<
7
;,(c) (x -1)^7 =^3 ·^5 ~n;^11 c^1312 (x -1)^6 for some c between
1 and x.
- f(x) = ex. For lxl < 2, Taylor's theorem says :Jc E (-2, 2) 3 IR6(x)I =
I
J<7
l(c)x11 = ~lxl^7 < e
2
27 =^128 e
2
7! 7! 7! 7! <^01877.. For lxl -<^1 ' Taylor's theoremsays :Jc E (-1, 1) 3 IR6(x)I = ~lxl^7 < ~~2^7 = fi < 0.00054.
- (a) We want IRn(x)I < 0.005 for all x E [-2, 2]. Since (n~l)! lxln+l <
e2.2n+1 (n+l)! , we can accomp l' is h t h' is b y ma k' mg e2.2n+1 (n+l)! < 0 005..
n 6 8 9 10
e2 2 n+l
(n+l)! .1877 .0104 .002^1 .00038