Calculus: f ( n ) = n for two distinct values of n implies that for some x, f ′ ( x ) = 1

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Calculus:   f  (  n  )  =  n for two distinct values of n implies that for some x,       f    ′    (  x  )  =  1

enreciarpv · Accepted Answer

Step 1Consider the function   g  :      R    →      R   defined by  g  (  x  )  =  f  (  x  )  −  xfor all   x  ∈      R   . Then, g is differentiable with       g    ′    (  x  )  =      f    ′    (  x  )  −  1 for all x, and by our assumption on f there exist two real numbers       x    1    &amp;lt;      x    2   such that   g  (      x    1    )  =  g  (      x    2    )  =  0 . By Rolle&#039;s theorem, there exists   y  ∈  (      x    1    ,      x    2    ) such that       g    ′    (  y  )  =  0 . That is, equivalently,       f    ′    (  y  )  =  1 .

Kody Arellano · Accepted Answer

Step 1By the Mean Value Theorem, there exists some c between n and m such that            f      (      n      )      −      f      (      m      )              n      −      m        =      f    ′    (  c  )

Calculus: f(n)=n for two distinct values of n implies that for some x, f′(x)=1

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