Use the intermediate value theorem to show that the equation, t a n ( x ) = 2 xhas an infinite amount of real solutions.So far I have used the IVT to show that for f ( x ) = t a n ( x ) in the interval ( − π 2 , π 2 ) there is a L between − ∞ and ∞ such that there is a value c where f ( c ) = LI also have shown that for g ( x ) = 2 x in the interval [ − π 2 , π 2 ] there is a K between − π and π such that there is a value d where f ( d ) = KWould it be correct to say that because each of these functions have a value for each point in the interval and that the range of f ( x ) consists of all reals that f ( x ) must intersect with g ( x ) or there is some value x in the interval such that 2 x = t a n ( x )? If this is correct then how would I extend this result to all intervals [ π ( n − 1 2 ) , π ( n + 1 2 ) ]?

Question

Use the intermediate value theorem to show that the equation,  t  a  n  (  x  )  =  2  xhas an infinite amount of real solutions.So far I have used the IVT to show that for   f  (  x  )  =  t  a  n  (  x  ) in the interval   (  −      π    2    ,      π    2    ) there is a   L between   −  ∞ and   ∞ such that there is a value   c where   f  (  c  )  =  LI also have shown that for   g  (  x  )  =  2  x in the interval   [  −      π    2    ,      π    2    ] there is a   K between   −  π and   π such that there is a value   d where   f  (  d  )  =  KWould it be correct to say that because each of these functions have a value for each point in the interval and that the range of   f  (  x  ) consists of all reals that   f  (  x  ) must intersect with   g  (  x  ) or there is some value   x in the interval such that   2  x  =  t  a  n  (  x  )? If this is correct then how would I extend this result to all intervals   [  π  (  n  −      1    2    )  ,  π  (  n  +      1    2    )  ]?

Clarissa Adkins · Accepted Answer

Consider   f  (  x  )  =  tan  ⁡  (  x  )  −  2  x. Then       f    ′    (  x  )  =            1                        cos          2                ⁡        (        x        )              −  2.On       π    4    ,      π    2   and   −      π    2    ,  −      π    4  ,       f    ′    (  x  )  &amp;gt;  0.Since       lim          x      →              π        2                    −                  f    ′    (  x  )  =  +  ∞  ,      lim          x      →              π        2                    −              f  (  x  )  =  +  ∞.Similarly since       lim          x      →      −              π        2                    +                  f    ′    (  x  )  =  −  ∞  ,      lim          x      →      −              π        2                    +              f  (  x  )  =  −  ∞.So   ∃      x    1    ∈  (  −      π    2    ,  −      π    4    ), such that,   f  (      x    1    )  &amp;lt;  0, and   ∃      x    2    ∈  (      π    4    ,      π    2    ), such that   f  (      x    2    )  &amp;gt;  0. By intermediate value theorem,   ∃      x    0    ,      x    1    &amp;lt;      x    0    &amp;lt;      x    2    ,  f  (      x    0    )  =  0 since   f  (  x  ) is continuous.Same reasoning can be applied to   −            π      2        +  k  π  ,            π      2        +  k  π,   k  ∈      Z  . So there are infinite real solution for   f  (  x  ).

Use the intermediate value theorem to show that the equation, tan(x)=2x has an infinite amount of real solutions.

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