If f ( x ) = x 2 and g ( x ) = x sin ⁡ ( x ) + cos ⁡ ( x ) then i have to find number of points(x) such that f(x)=g(x)I write h(x)=f(x)−g(x). So h ( x ) = x 2 − x sin ⁡ ( x ) − cos ⁡ ( x ) h ′ ( x ) = x ( 2 − cos ⁡ ( x ) ). Since 2 − cos ⁡ x is bounded so By taking limits as x goes to plus and minus infinity h′(x) goes to plus and minus infinity.which means h has root in between.

Question

If   f  (  x  )  =      x    2   and   g  (  x  )  =  x  sin  ⁡  (  x  )  +  cos  ⁡  (  x  ) then i have to find number of points(x) such that f(x)=g(x)I write h(x)=f(x)−g(x). So   h  (  x  )  =      x    2    −  x  sin  ⁡  (  x  )  −  cos  ⁡  (  x  )      h    ′    (  x  )  =  x  (  2  −  cos  ⁡  (  x  )  ). Since   2  −  cos  ⁡  x is bounded so By taking limits as x goes to plus and minus infinity h′(x) goes to plus and minus infinity.which means h has root in between.

radcas87gex5r · Accepted Answer

Note that   h  (  0  )  =  −  1  ≠  0. Therefore, in case of x is a root of h, then x&amp;gt;0 or x&amp;lt;0. Also,       lim          x      →      ±      ∞        h  (  x  )  =  ±  ∞ (Prove it!)...(*)If x&amp;gt;0, then h′(x)&amp;gt;0, so h is an increasing function on   (  0  ,  ∞  ), anb by continuity, also it is on   [  0  ,  ∞  ). But h(0)&amp;lt;0 and using (*) we conclude that there is a unique root of h in   (  0  ,  ∞  ) (Intermediate Value Theorem and monotonicity).Finally, an analogous argument shows that there is a unique root of h in   (  −  ∞  ,  0  ). Another way to conclude this is noting that h is an even function.Conclusion: there are exactly two roots of h.

If f ( x ) = x 2 </msup> and g ( x ) = x si

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