- Multiply by d. b< dXn-1 < XN-1 <1 d or dXN-1 2 b

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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show me the steps of determine red

dXN-k-eXn-I
Also,
AX, SA, BXN-k s B, CXn-i < C, DXN-c < D, anc
(E1)
...............
because each X; is a non-negative real less than or
equal to 1, i.e., X; e [, 1] C [0, 1].
Also,
Also,
AX, < A, BXN-k < B, CXN-1 < C, DXN-o < D, anc
<b
(E1)
..... ..........
because each X; is a non-negative real less than or
equal to 1, i.e., X; e , 1] C [0, 1].
So, specifically,
Xn-1 e 5, 1]
→ < XN-1 <1
Multiply by d.
b< dXn-1
or dXN-1 2 b
.... .... ...
Step 3
Now, consider the right side inequality.
XN-1 < 1
Multiply by
- e.
-eXN-1 2 -e
Using (E2) and (E3), we get
dXN-1 – eXN-1 2 b – e
Thus, if b + e, then b– e 0 and hence
1
dXN-1-eXN-1
b-e
Using inequalities (El) and (E4) in the equation (1),
we get
bXN-k
XN+1=AXN + BXN-k + CxN-1 + DXn-g +
dXN-k-eXp-i
b.
SA + B + C+D+
..... .... ....
b-e
Step 4
Transcribed Image Text:dXN-k-eXn-I Also, AX, SA, BXN-k s B, CXn-i < C, DXN-c < D, anc (E1) ............... because each X; is a non-negative real less than or equal to 1, i.e., X; e [, 1] C [0, 1]. Also, Also, AX, < A, BXN-k < B, CXN-1 < C, DXN-o < D, anc <b (E1) ..... .......... because each X; is a non-negative real less than or equal to 1, i.e., X; e , 1] C [0, 1]. So, specifically, Xn-1 e 5, 1] → < XN-1 <1 Multiply by d. b< dXn-1 or dXN-1 2 b .... .... ... Step 3 Now, consider the right side inequality. XN-1 < 1 Multiply by - e. -eXN-1 2 -e Using (E2) and (E3), we get dXN-1 – eXN-1 2 b – e Thus, if b + e, then b– e 0 and hence 1 dXN-1-eXN-1 b-e Using inequalities (El) and (E4) in the equation (1), we get bXN-k XN+1=AXN + BXN-k + CxN-1 + DXn-g + dXN-k-eXp-i b. SA + B + C+D+ ..... .... .... b-e Step 4
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