Let f : [ 0 , 1 ] → R continuous and let f ( [ 0 , 1 ] ) ⊂ [ 0 , 1 ] . By using intermediate value theorem, show that f has at least got one fixed point in [ 0 , 1 ] (aka there is one solution x ∈ [ 0 , 1 ] of the equation f ( x ) = x).I'm not sure at all if I did it right but I have taken the function: f ( x ) = xThen take the given intervals and insert for x the beginning of interval: f ( 0 ) = 0And the end of the interval: f ( 1 ) = 1 &gt; 0Because the function is continuous and because the equality sign changes to an inequality sign right after, the intermediate value theorem provides there must be at least one solution x 1 = [ 0 , 1 ].Did I do it correctly? Did I explain correctly?

Question

Let   f  :      [    0    ,    1    ]    →      R   continuous and let   f  (      [    0    ,    1    ]    )  ⊂      [    0    ,    1    ]  . By using intermediate value theorem, show that   f has at least got one fixed point in       [    0    ,    1    ]   (aka there is one solution   x  ∈      [    0    ,    1    ]   of the equation   f  (  x  )  =  x).I&#039;m not sure at all if I did it right but I have taken the function:  f  (  x  )  =  xThen take the given intervals and insert for   x the beginning of interval:  f  (  0  )  =  0And the end of the interval:  f  (  1  )  =  1  &amp;gt;  0Because the function is continuous and because the equality sign changes to an inequality sign right after, the intermediate value theorem provides there must be at least one solution       x          1        =  [  0  ,  1  ].Did I do it correctly? Did I explain correctly?

Derick Ortiz · Accepted Answer

Hint: Apply the intermediate value theorem to the function   g  (  x  )  =  f  (  x  )  −  x. Note that   g  (  0  )  ≥  0  ≥  g  (  1  ).

Let f:[0,1]->R continuous and let f([0,1])sub[0,1]. By using intermediate value theorem, show that f has at least got one fixed point in [0,1] (aka there is one solution x in [0,1] of the equation f(x)=x).

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