Show that tan ⁡ x − 2 x has a root in (0,1.4).I got: f ( x ) = tan ⁡ x − 2 x f ( 0 ) = 0, f ( 1.4 ) = − 2.775Does this mean there is no root in (0,1.4)? Because I got −2.775 and it did not fall in (0,1.4).

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Show that   tan  ⁡  x  −  2  x has a root in (0,1.4).I got:   f  (  x  )  =  tan  ⁡  x  −  2  x  f  (  0  )  =  0,   f  (  1.4  )  =  −  2.775Does this mean there is no root in (0,1.4)? Because I got −2.775 and it did not fall in (0,1.4).

Alison Mcgrath · Accepted Answer

If you put your calculator in radian mode, you&#039;ll see that   f actually takes on a positive value at 1.4. Now find an a in the interval where it takes a negative value. Since it is continuous on the closed interval from 0 to 1.4, then IVT says there is some   a in the interval (specifically,   c will be between a and 1.4) where the function is 0.Edit: It&#039;s worth noting that the function values found needn&#039;t fall in the given interval. For example, we can consider the continuous function  g  (  x  )  =  x  −  3on the interval [2,4]. Then   g  (  2  )  =  −  1 and   g  (  4  )  =  1  , so IVT says g has a zero in (2,4), meaning there is some   x  ∈  (  2  ,  4  ) such that   g  (  x  )  =  0. Of course,   0  ∉  (  2  ,  4  )--in fact, for any   x  ∈  [  2  ,  4  ]  , we have   g  (  x  )  ∉  [  2  ,  4  ]--but that has nothing to do with IVT.

Show that tanx−2x has a root in (0,1.4).

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