Let f be the function given by ƒ (z)=5 cos² () + In(z + 1) – 3 The derivative of f is given by f' (z) = -5 cos () sin() + What value of csatisfies the conclusion of the Mean Value Theorem applied to f on the interval [1, 4] ? 2.132 because f (2.132) F(4)–{(1) 3. 2.749 because f' (2.749) = F(4)–F(1) 3.042 because f' (3.042) = 0 3.252 because f' (3.252) f(1)+ f(4)
Let f be the function given by ƒ (z)=5 cos² () + In(z + 1) – 3 The derivative of f is given by f' (z) = -5 cos () sin() + What value of csatisfies the conclusion of the Mean Value Theorem applied to f on the interval [1, 4] ? 2.132 because f (2.132) F(4)–{(1) 3. 2.749 because f' (2.749) = F(4)–F(1) 3.042 because f' (3.042) = 0 3.252 because f' (3.252) f(1)+ f(4)
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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![Let f be the function given by f (z) =5 cos² () + In(z+1) – 3. The derivative of f is given by f' (z) = -5 cos () sin(3) + What value of c satisfies the conclusion of the Mean
Value Theorem applied to f on the interval [1, 4] ?
A
2.132 because f (2.132)
F(4)–F(1)
F(4)–{(1)
B
2.749 because f' (2.749)
3
3.042 because f' (3.042) = 0
f(1)+ f(4)
3.252 because f' (3.252)
23°F
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Transcribed Image Text:Let f be the function given by f (z) =5 cos² () + In(z+1) – 3. The derivative of f is given by f' (z) = -5 cos () sin(3) + What value of c satisfies the conclusion of the Mean
Value Theorem applied to f on the interval [1, 4] ?
A
2.132 because f (2.132)
F(4)–F(1)
F(4)–{(1)
B
2.749 because f' (2.749)
3
3.042 because f' (3.042) = 0
f(1)+ f(4)
3.252 because f' (3.252)
23°F
P Type here to search
中罗
Chp
fno 3
f12 E
home
end
lbrea
f5 CO
f6 CD
f7 CO
f3 O
f4 IOI
esc
f1
9
&
!
4
7
1
4
T.
Y
tab
Q
# M
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