df df-1 (c) Evaluate df-1 at x = a'and dx at x = f(a) to show that at these points dx dx dx) x+24 1. f(x): 1 a = ー x-=の %3D 1-x' (-)入

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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I need help to see if I am getting the right pattern to solve problem 1. I am having trouble with 1.c. please see if I made any mistakes and point out how I could prevent errs from occurring again. Thanks.
Classwork
For each function,
(a) Find f-1(x).
(b) Graph f and f-1 together.
df
df-1
at x = f(a) to show that at these points
(c) Evaluate
at x = a'and-
dx
%3D
dx
df-1
%3D
dx
1. f(x):
x+2
a =
1-x'
-x-1=の
(1-)チX
(a)
f(x) = X+2 then t() = X+2
x yt 2
R(-00, DUCI, 00)
→ y=の
1-y
%3D
1-X
%3D
(1-yXx)= y+ 2
#-メニ1)-2-x
2-X
y+ 2
y-X-2
ーX
%3D
リ-2-X+yx
-yx-y= 2-X-
Thomas' Calculus, 14e
P Pearson
Copyright © 2018, 2014, 2010 Pearson Education Inc.
Slide 16 of 95
Transcribed Image Text:Classwork For each function, (a) Find f-1(x). (b) Graph f and f-1 together. df df-1 at x = f(a) to show that at these points (c) Evaluate at x = a'and- dx %3D dx df-1 %3D dx 1. f(x): x+2 a = 1-x' -x-1=の (1-)チX (a) f(x) = X+2 then t() = X+2 x yt 2 R(-00, DUCI, 00) → y=の 1-y %3D 1-X %3D (1-yXx)= y+ 2 #-メニ1)-2-x 2-X y+ 2 y-X-2 ーX %3D リ-2-X+yx -yx-y= 2-X- Thomas' Calculus, 14e P Pearson Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 16 of 95
H.W. T.1
O Fraph f0 and f'6)
Tay
Find fo - FC+2)- a+2)
1-X
(i- Y = (x+2)-(-x)
Y (1-X) = (x+2)
ソ-yx = X+2
g= K-2
T+X
(x+2)
ソ-2
Y-2 = X(1)
X+yX
%3D
%3D
そu
「+リ
ソ-2 3 X
お「メ-2]1+x-1.X-2
仕X-X+2_
(1+x)*
3.
21
df
dx
%3D
[x+17, XP
3
X-2
ー1
= |+ X
3.
(+些 告
21
(1+x)-
1ーソーの
+x=o|RXキー
d (x+2)
rx-1リ7XP
|-x-Ex-2)_
1-x)
(1-x)
R(-00, 1)U(1,0)
1-X+x+2
(1-X)
3.
3.
(1-x)
Transcribed Image Text:H.W. T.1 O Fraph f0 and f'6) Tay Find fo - FC+2)- a+2) 1-X (i- Y = (x+2)-(-x) Y (1-X) = (x+2) ソ-yx = X+2 g= K-2 T+X (x+2) ソ-2 Y-2 = X(1) X+yX %3D %3D そu 「+リ ソ-2 3 X お「メ-2]1+x-1.X-2 仕X-X+2_ (1+x)* 3. 21 df dx %3D [x+17, XP 3 X-2 ー1 = |+ X 3. (+些 告 21 (1+x)- 1ーソーの +x=o|RXキー d (x+2) rx-1リ7XP |-x-Ex-2)_ 1-x) (1-x) R(-00, 1)U(1,0) 1-X+x+2 (1-X) 3. 3. (1-x)
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