Find the antiderivative of fLet S = { 0 } ∪ { 1 n : n ∈ N } and define f : R → R a) Prove that | f ( x ) − f ( y ) | ≤ | x − y | for any x , y ∈ R .b) Find the antiderivatives of fI obtained the following form for the function: f ( x ) = { − x , x ≤ 0 x − 1 n + 1 , 1 n + 1 ≤ x ≤ 1 2 ( 1 n + 1 + 1 n ) 1 n − x , 1 2 ( 1 n + 1 + 1 n ) &lt; x &lt; 1 n x − 1 , x ≥ 1 I then tried to prove the inequality from a) considering many cases for x and y, but there are many possibilities. For the antiderivatives, I tried to combine the antiderivatives for each of the 4 branches of the function, but again, the calculations are messy.

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Find the antiderivative of fLet   S  =  {  0  }  ∪  {      1    n    :  n  ∈      N    } and define   f  :      R    →      R  a) Prove that       |    f  (  x  )  −  f  (  y  )      |    ≤      |    x  −  y      |   for any   x  ,  y  ∈      R  .b) Find the antiderivatives of fI obtained the following form for the function:  f  (  x  )  =      {                            −          x          ,                                    x          ≤          0                                      x          −                      1                          n              +              1                                ,                                                1                          n              +              1                                ≤          x          ≤                      1            2                    (                      1                          n              +              1                                +                      1            n                    )                                                  1            n                    −          x          ,                                                1            2                    (                      1                          n              +              1                                +                      1            n                    )          &amp;lt;          x          &amp;lt;                      1            n                                                x          −          1          ,                                    x          ≥          1                        I then tried to prove the inequality from a) considering many cases for x and y, but there are many possibilities. For the antiderivatives, I tried to combine the antiderivatives for each of the 4 branches of the function, but again, the calculations are messy.

Leia Wiggins · Accepted Answer

Step 1Let S be a subset of       R  .1. For any   ϵ  &amp;gt;  0 there is       s    ′    ∈  S such that       |    x  −      s    ′        |    ≤  f  (  x  )  +  ϵ. We know that   f  (  y  )  ≤      |    y  −      s    ′        |    =      |    y  −  x  +  x  −      s    ′        |    ≤      |    y  −  x      |    +  f  (  x  )  +  ϵ.Since this holds for any   ϵ  &amp;gt;  0 we have   f  (  y  )  ≤  f  (  x  )  +      |    y  −  x      |  .Step 22. For any   δ  &amp;gt;  0 there is       t    ′    ∈  S such that       |    y  −      t    ′        |    ≤  f  (  y  )  +  δ. Analogously,   f  (  x  )  ≤      |    x  −      t    ′        |    =      |    x  −  y  +  y  −      t    ′        |    ≤      |    x  −  y      |    +  f  (  y  )  +  δ.Since this holds for any   δ  &amp;gt;  0 we have   f  (  x  )  ≤  f  (  y  )  +      |    y  −  x      |  .Step 33. Thus we have   f  (  x  )  −  f  (  y  )  ≤      |    x  −  y      |      f  (  y  )  −  f  (  x  )  ≤      |    y  −  x      |    .Thus       |    f  (  y  )  −  f  (  x  )      |    ≤      |    y  −  x      |  .No let S be defined as in your example and you are done.

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New Questions in Calculus 1