How would I solve the following two questions.Using the intermediate value theorem to show that there is a solution of the equation s i n 2 x 2 − x + 1 = 0 in the interval [ 0 , p i ]I showed by the IVT there is a c in [ 0 , p i ] give that c is zero because 0 &lt; 1 0 &gt; − p i + 1 but I am not sure if it did this correctly.2.My second question asked me to sketch a function that satisfies the following conditions or prove that it is impossible f ( x ) is a continous function on [0,2] its minimum value is −3 but it does not have a max value.I said this is impossible because if f ( x ) is continous on a bounded interval it must have both a max and min value.

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How would I solve the following two questions.Using the intermediate value theorem to show that there is a solution of the equation             s      i              n        2            x        2    −  x  +  1  =  0 in the interval   [  0  ,  p  i  ]I showed by the IVT there is a c in   [  0  ,  p  i  ] give that c is zero because   0  &amp;lt;  1   0  &amp;gt;  −  p  i  +  1 but I am not sure if it did this correctly.2.My second question asked me to sketch a function that satisfies the following conditions or prove that it is impossible   f  (  x  ) is a continous function on [0,2] its minimum value is −3 but it does not have a max value.I said this is impossible because if   f  (  x  ) is continous on a bounded interval it must have both a max and min value.

Kingston Oliver · Accepted Answer

Using the intermediate value theorem to show that there is a solution of the equation                               sin          2                ⁡        x            2        −  x  +  1  =  0 in the interval   [  0  ,  π  ]&quot;I showed by the IVT there is a c in   [  0  ,  π  ] give that c is zero because   0  &amp;lt;  1,   0  &amp;gt;  −  p  i  +  1 but I am not sure if it did this correctly.&quot;Your process in answering this is just fine: just clarify the details:Let   f  (  x  )  =                              sin          2                ⁡        x            2        −  x  +  1.Show the Intermediate Value theorem does, in fact, apply on the given interval   [  0  ,  π  ]: Then   f  (  0  )  &amp;gt;  0 and   f  (  π  )  &amp;lt;  0, using your computations...etc., including the details/justifications, you posted.The second part looks just fine, as it is, as you explained exactly why it is not possible.

Dayana Doyle · Accepted Answer

because   f  (  0  )  =  0  &amp;lt;  1 and because   f  (  p  i  )  =  −  p  i  +  1  &amp;lt;  0 i guess

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