Question

Show that there is a number c. with 0 Sc51. such that/100 f(x)=6xcosx

a. We have that /(0)=-10 and/(1)6-cos1> 0 and that / is periodic. Thus, by the Intermediate Value
b. Theorem applied to 0, there is a number c in [0, 1] such that /(c)=0. We have that/(0)10 and f(1)=6-cos1 > 0 and that is continuous. Thus, by the Intermediate Value Theorem applied to k = 0, there is a number c in [0, 1] such that (c)==0
c. We have that /(0) 10and/(1)6-cos1 > 0 and that/is continuous. Thus, by the Intermediate Value Theorem applied to 40, there is a number c in [0, 1] such that /(c)==0.
d. We have that /(0)-10and/(1)=6>0 ae that is continuous. Thus, by the intermediate Valve Theorem
applied to 0. there is a number c in [0, 1] such that /(c) We have that (0)1> 0 and/(1)=6> 0 and that is periodic. Thus, by the Intermediate Value Theorem applied to 40. there is a number c in [0, 1] such that (c) == 0.

Answer 1

The question essentially asks to prove that a continuous and periodic function, with specific values at the endpoints of an interval, has a root within that interval according to the Intermediate Value Theorem.

Step-by-step explanation:

The question involves applying the Intermediate Value Theorem to a continuous function to find a number c for which f(c) is equal to 0 within a specified interval. Specifically, it is suggested that the function in question has particular values f(0) = -10 and f(1) = 6 - cos(1) > 0. The continuity of the function ensures that if it changes sign over the interval [0, 1], there must be at least one point within that interval where the function is zero. Since the function is periodic and continuous, the Intermediate Value Theorem guarantees the existence of such a point, labeled c, where 0 ≤ c ≤ 1 and f(c) = 0.