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 continuous ✔✔✔ on the interval [1, 2] since fis the composite of the sine function and the cubing function, both of which are continuous ✔✔✔ on R. The zeros of sin(x) are at x - 0 <1 <# 3³ <2m <8 < 3m. Applying the intermediate value theorem and taking the pertinent cube roots these values in the inequality, we have the following. [call this value A] < 2 1<√ 01<√3 [call this value A] 0, then substituting the value A found above and simplifying, f(A)=sin( such that f(c) f(0) = Thus, f has at least two x-intercepts in (1, 2). Need Help? Read It x forn in Z, so we note that < 0, and f(2)=sin(8) > 0. Applying the intermediate value theorem on [1, A] and then on [A, 2], we see there are numbers c and d in (1, A) and (A, 2),
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 continuous ✔✔✔ on the interval [1, 2] since fis the composite of the sine function and the cubing function, both of which are continuous ✔✔✔ on R. The zeros of sin(x) are at x - 0 <1 <# 3³ <2m <8 < 3m. Applying the intermediate value theorem and taking the pertinent cube roots these values in the inequality, we have the following. [call this value A] < 2 1<√ 01<√3 [call this value A] 0, then substituting the value A found above and simplifying, f(A)=sin( such that f(c) f(0) = Thus, f has at least two x-intercepts in (1, 2). Need Help? Read It x forn in Z, so we note that < 0, and f(2)=sin(8) > 0. Applying the intermediate value theorem on [1, A] and then on [A, 2], we see there are numbers c and d in (1, A) and (A, 2),
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 f is continuous
</
0 <1 <n<
Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following.
1<
[call this value A] < 2
0 1 <√√37 [call this value A] <√√/2
O 1 <3 [call this value A] < 2
[call this value A] <√√2
< 2π < 8 < 3π.
01<
Need Help?
✓ on the interval [1, 2] since fis the composite of the sine function and the cubing function, both of which are [continuous ✓on R. The zeros of sin(x) are at x =
Now f(1) = sin(1) > 0, then substituting the value A found above and simplifying, f(A) = sin
such that f(c) = f(d) =
Thus, f has at least two x-intercepts in (1, 2).
Read It
X
X
for n in Z, so we note that
< 0, and f(2)=sin(8) > 0. Applying the intermediate value theorem on [1, A] and then on [A, 2], we see there are numbers c and d in (1, A) and (A, 2),](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2028644b-9cb0-43c2-8826-9838d27acbaa%2F36b38bce-bc8a-46fd-9134-3cd67d2e1259%2Fnjihfx_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 f is continuous
</
0 <1 <n<
Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following.
1<
[call this value A] < 2
0 1 <√√37 [call this value A] <√√/2
O 1 <3 [call this value A] < 2
[call this value A] <√√2
< 2π < 8 < 3π.
01<
Need Help?
✓ on the interval [1, 2] since fis the composite of the sine function and the cubing function, both of which are [continuous ✓on R. The zeros of sin(x) are at x =
Now f(1) = sin(1) > 0, then substituting the value A found above and simplifying, f(A) = sin
such that f(c) = f(d) =
Thus, f has at least two x-intercepts in (1, 2).
Read It
X
X
for n in Z, so we note that
< 0, and f(2)=sin(8) > 0. Applying the intermediate value theorem on [1, A] and then on [A, 2], we see there are numbers c and d in (1, A) and (A, 2),
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