Constructing a function that is continuous and has a max on an open interval, but is not necessarily increasing immediately to the left of the max.Suppose F(x) is continuous on some open interval I and c is a maximum point inside this interval. Is it true that f(x) must be increasing immediately to the left of c and decreasing immediately to the right of c? Proof or counterexample. (Note: A constant function is considered to be both increasing and decreasing.)

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cydostwng6c · Accepted Answer

Step 1  f  (  x  )  =      x    2        (    −    2    +    sin    ⁡          1      x        )  when   x  ≠  0, and   f  (  0  )  =  0.By squeezing, you can see that this is continuous at 0Step 2Since   sin  ⁡  (  anything  )  ≤  1, we have   −  2  +  sin  ⁡            1      x        ≤  −  1, so   f  (  x  )  ≤  −      x    2  , but   f  (  x  )  =  0. So it has a global maximum at 0.Now work on showing that no matter how close to 0 you get, there are places where the derivative has the wrong sign for f to satisfy the false conjecture.

atestiguoki · Accepted Answer

Explanation:Define   f  :      R    →      R        {                            f          (          x          )                          :=          −                      |                    x          sin          ⁡                      1            x                                |                    ,                          if           x          ≠          0          ,                                              =          0          ,                          otherwise          .

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