f is continuous function in [ 0 , 2 ] and f ( 2 ) = 1proof that there is a point x in [ 0 , 2 ] so that: f ( x ) = 1 x nobody in my class solved this exercise. our teacher give us a hint: p ( x ) = x ∗ f ( x ) and to do Intermediate value theorem on this function. so I tried: p ( 0 ) = 0 ∗ f ( 0 ) = 0 , P ( 2 ) = 2 ∗ f ( 2 ) = 2 p ( 0 ) &lt; h &lt; p ( 2 ) by Intermediate value theorem, there is a point c so that a &lt; c &lt; b and f ( c ) = hplease solve

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f is continuous function in   [  0  ,  2  ] and   f  (  2  )  =  1proof that there is a point   x in   [  0  ,  2  ] so that:  f  (  x  )  =      1    x  nobody in my class solved this exercise. our teacher give us a hint:   p  (  x  )  =  x  ∗  f  (  x  ) and to do Intermediate value theorem on this function. so I tried:  p  (  0  )  =  0  ∗  f  (  0  )  =  0  ,  P  (  2  )  =  2  ∗  f  (  2  )  =  2  p  (  0  )  &amp;lt;  h  &amp;lt;  p  (  2  ) by Intermediate value theorem, there is a point   c so that   a  &amp;lt;  c  &amp;lt;  b and   f  (  c  )  =  hplease solve

saniraore · Accepted Answer

You seem to be almost done. You know that   p  (  0  )  =  0,   p  (  2  )  =  2 and that   p is continuous. You want to find an   x  ∈  [  0  ,  2  ] such that   f  (  x  )  =  1      /    x, in other words   p  (  x  )  =  1. Thus you can set   h  =  1 and you know there exists a   c  ∈  [  0  ,  2  ] such that   p  (  c  )  =  1. This in turn means   f  (  c  )  =  1      /    c, so you&#039;re done.

sponsorjewk · Accepted Answer

Hint: If   p  (  x  )  =  h, then  f  (  x  )  =      h    x  You&#039;re free to choose   h to be anything between 0 and 2, and there&#039;s a particularly salient choice of   h.

f is continuous function in [0,2] and f(2)=1 proof that there is a point x in [0,2] so that: f(x)=1/x

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