Consider the function f ( x ) = sup { x 2 , cos ⁡ x }, x ∈ [ 0 , π 2 ]. Show that there exists a point x 0 ∈ [ 0 , π 2 ] such that f ( x 0 ) = min x ∈ [ 0 , π 2 ] f ( x ). Also show that cos ⁡ x 0 = x 0 2 .The first part is obvious due to continuity of f in [ 0 , π 2 ]. Second part is also obvious from the graph of f. But how to mathematically prove that cos ⁡ x 0 = x 0 2 ?I was trying to show it using the expression of f f ( x ) = 1 2 { x 2 + cos ⁡ x + | x 2 − cos ⁡ x | }. Can anybody please help me out?

Question

Consider the function   f  (  x  )  =  sup  {      x    2    ,  cos  ⁡  x  },   x  ∈  [  0  ,            π      2        ]. Show that there exists a point       x    0    ∈  [  0  ,            π      2        ] such that   f  (      x    0    )  =      min          x      ∈      [      0      ,                        π          2                    ]        f  (  x  ). Also show that   cos  ⁡      x    0    =      x    0    2  .The first part is obvious due to continuity of f in   [  0  ,            π      2        ]. Second part is also obvious from the graph of f. But how to mathematically prove that   cos  ⁡      x    0    =      x    0    2  ?I was trying to show it using the expression of f   f  (  x  )  =            1      2        {      x    2    +  cos  ⁡  x  +      |        x    2    −  cos  ⁡  x      |    }. Can anybody please help me out?

Payten Daniels · Accepted Answer

Step 1If not, then let   c  o  s         x    0    &amp;gt;  (      x    o        )    2  . Then by the definition of f,   f  (      x    0    )  =  c  o  s         x    0      ⟹       c  o  s         x    0   is the minimum value when   x  ∈  [  0  ,      π    2    ], which is a contradiction to the fact that   c  o  s         x    0    &amp;gt;  (      x    0        )    2   as it turns out to be the minimum value.Step 2Similarly if   (      x    0        )    2    &amp;gt;  c  o  s         x    o   then again we have a contradiction.

Raelynn Nolan · Accepted Answer

Step 1Consider an       x    ∗    ∈            [        0  ,            π      2                  ]      . If   cos  ⁡      x    ∗    &amp;gt;      x    ∗    2   then (a)       x    ∗    ≠            π      2      , and (b)   f  (  x  )  =  cos  ⁡  x in a full neighborhood of       x    ∗  . As cos is strictly decreasing in             [        0  ,            π      2                  ]       there are points x immediately to the right of       x    ∗   where   f  (  x  )  &amp;lt;  f  (      x    ∗    ). In a similar way one argues when   cos  ⁡      x    ∗    &amp;lt;      x    ∗    2  . In this case       x    ∗    ≠  0, and there are points x immediately to the left of       x    ∗   where   f  (  x  )  &amp;lt;  f  (      x    ∗    ).Step 2This allows to conclude that at the point       x    0   where f is minimal one necessarily has   cos  ⁡      x    0    =      x    0    2  .Note that we have used the monotonicity of cos and the square function in the given interval. Without this assumption it would be easy to give counterexamples.

Consider the function f(x)=sup{x^2, cos x}, x in [0, pi/2]. Show that there exists a point x_0 in [0, pi/2] such that f(x_0)=min_{x in[0, pi/2] f(x). Also show that cos x_0 = x_0^2.

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