A continuous function y = f(x) is known to be negative at x = 5 and positive at x = 8. Why does the equation f(x) = 0 have at least one solution between x = 5 and x = 8? Illustrate with a sketch. Why does the equation f(x) = 0 have at least one solution between x = 5 and x = 8? O A. f(x)=0 has at least one solution between x = 5 and x = 8 because f(x) must pass through all values between f(5) and f(8), regardless of whether f is continuous. O B. f(x) = 0 has at least one solution between x = 5 and x = 8 because f is a continuous function on the closed interval [5, 8], and if yo is any value between f(5) and f(8), then yo = f(c) for some c in [5, 8]. O c. f(x)=0 has at least one solution between x = 5 and x = 8 because all continuous functions have at least one zero over any nonempty closed interval. Choose a graph below that illustrates the situation. O A. 2- Q x Q ✔ OB. Q OC. -2- Q O D. 2-
A continuous function y = f(x) is known to be negative at x = 5 and positive at x = 8. Why does the equation f(x) = 0 have at least one solution between x = 5 and x = 8? Illustrate with a sketch. Why does the equation f(x) = 0 have at least one solution between x = 5 and x = 8? O A. f(x)=0 has at least one solution between x = 5 and x = 8 because f(x) must pass through all values between f(5) and f(8), regardless of whether f is continuous. O B. f(x) = 0 has at least one solution between x = 5 and x = 8 because f is a continuous function on the closed interval [5, 8], and if yo is any value between f(5) and f(8), then yo = f(c) for some c in [5, 8]. O c. f(x)=0 has at least one solution between x = 5 and x = 8 because all continuous functions have at least one zero over any nonempty closed interval. Choose a graph below that illustrates the situation. O A. 2- Q x Q ✔ OB. Q OC. -2- Q O D. 2-
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 92E
Related questions
Question
![A continuous function y = f(x) is known to be negative at x = 5 and positive at x = 8. Why does the equation f(x) = 0 have at least one solution between x = 5 and x = 8? Illustrate with a sketch.
Why does the equation f(x) = 0 have at least one solution between x = 5 and x = 8?
A. f(x) = 0 has at least one solution between x = 5 and x = 8 because f(x) must pass through all values between f(5) and f(8), regardless of whether f is continuous.
B. f(x) = 0 has at least one solution between x = 5 and x = 8 because f is a continuous function on the closed interval [5, 8], and if y is any value between f(5) and f(8), then yo = f(c) for some c in [5, 8].
C. f(x) = 0 has at least one solution between x = 5 and x = 8 because all continuous functions have at least one zero over any nonempty closed interval.
Choose a graph below that illustrates the situation.
A.
AY
2-
0
ST
▬▬
X
LO
B.
Ay
2-
0
=
L
C.
Ay
2-
10
D.
2-
-2-
ENE
=
10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84504046-073c-4202-8178-d38af176ff2d%2Fb50fe4c7-c0c1-4bc3-99c8-7c955335f840%2Fyc3l05b_processed.png&w=3840&q=75)
Transcribed Image Text:A continuous function y = f(x) is known to be negative at x = 5 and positive at x = 8. Why does the equation f(x) = 0 have at least one solution between x = 5 and x = 8? Illustrate with a sketch.
Why does the equation f(x) = 0 have at least one solution between x = 5 and x = 8?
A. f(x) = 0 has at least one solution between x = 5 and x = 8 because f(x) must pass through all values between f(5) and f(8), regardless of whether f is continuous.
B. f(x) = 0 has at least one solution between x = 5 and x = 8 because f is a continuous function on the closed interval [5, 8], and if y is any value between f(5) and f(8), then yo = f(c) for some c in [5, 8].
C. f(x) = 0 has at least one solution between x = 5 and x = 8 because all continuous functions have at least one zero over any nonempty closed interval.
Choose a graph below that illustrates the situation.
A.
AY
2-
0
ST
▬▬
X
LO
B.
Ay
2-
0
=
L
C.
Ay
2-
10
D.
2-
-2-
ENE
=
10
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