Hello, Please help with the following: Show that equation 5x - 2 sinx = 0 has ex

Question

Hello,

Please help with the following:

Show that equation 5x - 2 sinx = 0 has exactly one root with the following two steps:

(a)  Use the Intermediate Value Theorem to show that the equation has a root of between 0 and pi.

(b)  Assumint the equation has at least two real roots, derive the a contradiction using Rolle's Theorem.

(Book Answer for (a) Letting f(x)= 5x-2sinx, f takes the value 0 somewhere in [0, pi]

and for (b)  If there are two roots, then f(a) = f(b) = 0 and f'(c) = 0 or 5 + cosc = 0 which can't be)

Please show step by step solution.  If (b) is too complex, I'd still appreciate some explanation for (a).

Thanks

Explanation / Answer

(a) Statement of the Intermediate Value Theorem: if f is a real-valued continuous function on the interval [a, b], and u is a number between f(a) and f(b), then there is a c in (a, b) such that f(c) = u.

Solution: In this case, f(0)=0, and f(pi)=5pi. Hence, there is a c such that f(c)=0 by the IVT. (In other words, c is a root).

(b) Statement of Rolle's Theorem: If f is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists a c in the open interval (a, b) such that f'(c)=0.

Solution: Assume, for the sake of contradiction, that there are two roots. Let these roots be at the points a and b. Then f(a)=f(b)=0. By Rolle's theorem, there must exist a c such that f'(c)=0. But the derivative of your function gives f'(x)=5-2cos(x).

Letting f'(x)=0, we want to solve 5+2cos(x) = 0. This never happens, since there is no such value x for which this is true within the interval. Hence, we have a contradiction, and there must be only one root.

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