Let f be continuous on [ 0 , 1 ] and such that f ( 0 ) &lt; 0 and f ( 1 ) &gt; 1. Prove that there exists c ∈ ( 0 , 1 ) such that f ( c ) = c 4

Question

Let   f be continuous on   [  0  ,  1  ] and such that   f  (  0  )  &amp;lt;  0 and   f  (  1  )  &amp;gt;  1. Prove that there exists   c  ∈  (  0  ,  1  ) such that   f  (  c  )  =      c    4

Haylie Campbell · Accepted Answer

If we let   g  (  x  )  =  f  (  x  )  −      x    4  , then we are given   g  (  0  )  =  f  (  0  )  −  0  &amp;lt;  0 and   g  (  1  )  =  f  (  1  )  −  1  &amp;gt;  0. Since   f is continuous and       x    4   is continuous, their difference is, and   g is defined on the interval   [  0  ,  1  ], so we can apply the Intermediate Value Theorem to   g. In particular, since   g  (  0  )  &amp;lt;  0  &amp;lt;  g  (  1  ), there exists   c  ∈  (  0  ,  1  ) such that   g  (  c  )  =  0, so   f  (  c  )  −      c    4    =  0, or   f  (  c  )  =      c    4   as we desiblack.

seguitzla · Accepted Answer

consider  g  (  x  )  =  f  (  x  )  −      x    4    g  (  x  ) is continuous from   [  0  ,  1  ] since both   f  (  x  ) and       x    4   continuous on   [  0  ,  1  ].Now,   g  (  0  )  =  f  (  0  )  −      0    4    =  a  &amp;lt;  0 (since   f  (  0  )  &amp;lt;  0 is given)  g  (  1  )  =  f  (  1  )  −      1    4    =  b  &amp;gt;  0 (since   f  (  1  )  &amp;gt;  1 is given)  g  (  x  ) is continuous, so by Intermediate Value Theorem, for every   d between   g  (  0  ) and   g  (  1  )   ∃    c  ∈  (  0  ,  1  ) such that   g  (  c  )  =  dChoose   d  =  0 so we have   g  (  c  )  =  0 or   g  (  c  )  =  f  (  c  )  −      c    4    =  0 which implies  f  (  c  )  =      c    4      f  o  r  c  ∈  (  0  ,  1  )

Let f be continuous on [0,1] and such that f(0)<0 and f(1)>1. Prove that there exists c in (0,1) such that f(c)=c^4

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