MHB Lamps's question at Yahoo Answers about the Intermediate Value Theorem

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The discussion addresses the application of the Intermediate Value Theorem (IVT) to find intervals containing roots for two equations. For the first equation, sin(x) = 6x + 5, it is established that f(x) = sin x - 6x - 5 is continuous, with f(-1) > 0 and f(0) < 0, indicating a root exists in the interval (-1, 0). For the second equation, ln(x) + x^2 = 3, g(x) = ln x + x^2 - 3 is continuous on (0, +∞), with g(1) < 0 and g(2) > 0, confirming a root in the interval (1, 2). The responses effectively demonstrate the use of the IVT to locate roots within specified intervals. This method highlights the importance of continuity in applying the theorem to find solutions.
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Here is the question:

use the IVT to find the an interval of length one that contains a root of the equation
a) sin(x) = 6x + 5

b) ln(x) + x^2 = 3

Here is a link to the question:

Intermediate value theorem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello lamp,

(a) Denote f(x)=\sin x-6x-5. Clearly, f in continuos in \mathbb{R}. We have:

f(-1)=\sin (-1)+6-5=1-\sin 1&gt;0,\quad f(0)=-5&lt;0

Then, 0\in (f(0),f(-1)) and according to the Intermediate Value Theorem there exists a\in (-1,0) such that f(a)=0 or equivalently \sin a=6a+5

(b) Now, denote g(x)=\ln x+x^2-3. Clearly, f in continuos in (0,+\infty). We have:

g(1)=-2&lt;0,\quad g(2)=\ln 2+1&gt;0

and we can reason as in (a).
 
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