Show that the following equation has exactly one real root: 3 x + 2 cos ⁡ x + 5 = 0So I think the way to approach this problem is to use a combination of the intermediate value theorem and the mean value theorem.But since we are not given a specific interval, I'm not entirely sure what values of f ( a ) and f ( b ) to choose. Am I suppose to choose f ( 0 ) and f ( 1 ) and see that 0 lies in between them. If so, how exactly can we proceed with the mean value theorem after using the intermediate value theorem. Do we have to check for continuity and the number c, or is there some other way?

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Show that the following equation has exactly one real root:  3  x  +  2  cos  ⁡  x  +  5  =  0So I think the way to approach this problem is to use a combination of the intermediate value theorem and the mean value theorem.But since we are not given a specific interval, I&#039;m not entirely sure what values of   f  (  a  ) and   f  (  b  ) to choose. Am I suppose to choose   f  (  0  ) and   f  (  1  ) and see that 0 lies in between them. If so, how exactly can we proceed with the mean value theorem after using the intermediate value theorem. Do we have to check for continuity and the number c, or is there some other way?

faux0101d · Accepted Answer

Let   f  (  x  )  =  3  x  +  2  cos  ⁡  x  +  5. Then       f    ′    (  x  )  =  3  −  2  sin  ⁡  x  &amp;gt;  0. Thus   f is strictly increasing. Then graph of   f can intersect the   x-axis at most once.Note   f is continuous and       lim          x      →      −      ∞        f  (  x  )  =  −  ∞ and   f  (  0  )  =  5  &amp;gt;  0, hence it&#039;s graph must intersect the   x-axis at least once.

Nikolai Decker · Accepted Answer

Guide:Let   f  (  x  )  =  3  x  +  2  cos  ⁡  x  +  5,      f    ′    (  x  )  =  3  −  2  sin  ⁡  (  x  )  &amp;gt;  0Also compute limit of   x when   x tends to   −  ∞ and also   ∞.

Show that the following equation has exactly one real root: 3x+2cosx+5=0

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