Applying IV theorem to practical problemsProve using the IVT: For all positive integers k, c o s x = x k has a solutionI know that the IV states at an interval [a,b] some value c is bound to be hit atleast once, assuming that [a,b] is continuous(Please improve my explanation where needed).So, using this logic, assuming that 0 = x k − c o s x, I have to find values of x where this would be true? f ( 0 ) = 0 − 1 f ( π / 2 ) = ( π / 2 ) k − 0 f ( π / 2 ) &gt; 0By the IVT, there should be a value C inbetween [ 0 , π / 2 ] if I am correct?Please correct me where possible. I think I understand the theorem but applying it to a question like this makes me believe otherwise.

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Applying IV theorem to practical problemsProve using the IVT: For all positive integers k,   c  o  s  x  =      x    k   has a solutionI know that the IV states at an interval [a,b] some value c is bound to be hit atleast once, assuming that [a,b] is continuous(Please improve my explanation where needed).So, using this logic, assuming that   0  =      x    k    −  c  o  s  x, I have to find values of x where this would be true?  f  (  0  )  =  0  −  1  f  (      π        /    2  )  =  (      π        /    2      )    k    −  0  f  (      π        /    2  )  &amp;gt;  0By the IVT, there should be a value C inbetween   [  0  ,      π        /    2  ] if I am correct?Please correct me where possible. I think I understand the theorem but applying it to a question like this makes me believe otherwise.

Marnovdk · Accepted Answer

Step 1Your interpretation is correct. As to writing style, you should give a definition of f(x) before discussing its properties.Step 2The derivative of   f  (  x  )  =      x    k    −  cos  ⁡  x is       f    ′    (  x  )  =  k      x          k      −      1        +  sin  ⁡  x  ,, which is positive for   x  &amp;gt;  0 for any positive k. So f(x) is strictly increasing, and continuous, for   x  ≥  0 and   k  &amp;gt;  0.. And so, since   f  (  0  )  &amp;lt;  0  &amp;lt;  f  (  π      /    2  )  , there is a unique   x  ∈  (  0  ,  π      /    2  ) such that   f  (  x  )  =  0..

Prove using the IVT: For all positive integers k, cos x=x^k.

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