Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval. y=sin(x³), (1, 2) Let f(x)=sin(x³). Then fis ---Select--- on the interval [1, 2] since f is the composite of the sine function and the cubing function, both of which are ---Select-- Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following. 01<√ [call this value A] < 2 ✓on R. The zeros sin(x) are at x- for n in Z, so we note that 0 <1 << < 2m < 8 < 3m.
Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval. y=sin(x³), (1, 2) Let f(x)=sin(x³). Then fis ---Select--- on the interval [1, 2] since f is the composite of the sine function and the cubing function, both of which are ---Select-- Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following. 01<√ [call this value A] < 2 ✓on R. The zeros sin(x) are at x- for n in Z, so we note that 0 <1 << < 2m < 8 < 3m.
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 35E
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![Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval.
y = sin(x³), (1, 2)
Let f(x) = sin(x³). Then fis ---Select--- ✓on the interval [1, 2] since f is the composite of the sine function and the cubing function, both of which are ---Select--- ✓on R. The zeros of sin(x) are at x =
Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following.
0 1 <i
[call this value A] < 2
0 1 <√√/37 [call this value A] <√√/2
0 1 <3 [call this value A] < 2
3
<√√/=/ T
0 1
[call this value A]<√√2
for n in Z, so we note that 0 <1 << < 3/7 < 2π < 8 < 3π.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2028644b-9cb0-43c2-8826-9838d27acbaa%2F5c5ccfa0-8a67-4c2d-bd60-5fcf713c171d%2Fcaozb58_processed.png&w=3840&q=75)
Transcribed Image Text:Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval.
y = sin(x³), (1, 2)
Let f(x) = sin(x³). Then fis ---Select--- ✓on the interval [1, 2] since f is the composite of the sine function and the cubing function, both of which are ---Select--- ✓on R. The zeros of sin(x) are at x =
Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following.
0 1 <i
[call this value A] < 2
0 1 <√√/37 [call this value A] <√√/2
0 1 <3 [call this value A] < 2
3
<√√/=/ T
0 1
[call this value A]<√√2
for n in Z, so we note that 0 <1 << < 3/7 < 2π < 8 < 3π.
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