Let f ( x ) = x 2 sin ⁡ ( 1 x ) for x ≠ 0 and f ( 0 ) = 0.Show that 0 is a critical point of f that is not a local maximum nor a local minimum nor an inflection point.

Question

Let  f  (  x  )  =      x    2    sin  ⁡      (          1      x        )  for   x  ≠  0 and   f  (  0  )  =  0.Show that   0 is a critical point of   f that is not a local maximum nor a local minimum nor an inflection point.

Toby Barron · Accepted Answer

Prove it is a critical point (that the derivative exists and it is   0).Prove it is neither a maximum or a minimum by proving that for any   ε  &amp;gt;  0 there are   x and   y,   0  &amp;lt;  x  &amp;lt;  ε  ,  0  &amp;lt;  y  &amp;lt;  ε such as   f  (  x  )  &amp;gt;  f  (  0  ) and   f  (  y  )  &amp;lt;  f  (  0  ) (and   f  (  −  x  )  &amp;lt;  f  (  0  ) and   f  (  −  y  )  &amp;gt;  f  (  0  )) so there is no neighborhood in which   0 is maximum or minimum.This also implies why it is not an inflection point.

Let f(x)=x^(2)sin(1/x) for x!=0 and f(0)=0 .

Answered question

Answer & Explanation

New Questions in Differential Calculus