Antiderivative Theory ProblemA function f is differentiable over its domain and has the following properties:1. f ( x + y ) = f ( x ) + f ( y ) 1 − f ( x ) f ( y ) .2. lim h → 0 f ( h ) = 03. lim h → 0 f ( h ) / h = 1.i) Show that f ( 0 ) = 0.ii) show that f ′ ( x ) = 1 + [ f ( x ) ] 2 by using the def of derivatives Show how the above properties are involved.iii) find f(x) by finding the antiderivative. Use the boundary condition from part (i).So basically I think I found out how to do part 1 because if x + y = 0 then the top part of the fraction will always have to be zero.part 2 and 3 are giving me trouble. The definition is the limit ( f ( x + h ) − f ( x ) ) / h.So I can set x + y = h and make the numerator equal to f(h)?

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Antiderivative Theory ProblemA function f is differentiable over its domain and has the following properties:1.       f    (    x    +    y    )    =                  f        (        x        )        +        f        (        y        )                    1        −        f        (        x        )        f        (        y        )            .2.       lim          h      →      0        f  (  h  )  =  03.       lim          h      →      0        f  (  h  )      /    h  =  1.i) Show that   f  (  0  )  =  0.ii) show that       f    ′    (  x  )  =  1  +  [  f  (  x  )      ]    2   by using the def of derivatives Show how the above properties are involved.iii) find f(x) by finding the antiderivative. Use the boundary condition from part (i).So basically I think I found out how to do part 1 because if   x  +  y  =  0 then the top part of the fraction will always have to be zero.part 2 and 3 are giving me trouble. The definition is the limit   (  f  (  x  +  h  )  −  f  (  x  )  )      /    h.So I can set   x  +  y  =  h and make the numerator equal to f(h)?

Calvin Oneill · Accepted Answer

Step 1  f  (  x  )  =            f      (      x      )      −              f        2            (      x      )      f      (      h      )              1      −      f      (      x      )      f      (      h      )      Now I&#039;ll look at 3). In order to solve the ODE:            d      f              d      x        =  1  +      f    2  Step 2We simply separate the parts out to get  ∫            d      f              1      +              f        2              =  ∫  d  xThus   arctan  ⁡      f    =  x  +  C    ⟺    f  =  tan  ⁡      (    x    +    C    )  Now we see from part a) that   f  (  0  )  =  0 so  0  =  tan  ⁡  (  0  +  C  )    ⟺    C  =  arctan  ⁡      0    =  0and we get our solution of   f  (  x  )  =  tan  ⁡  (  x  )

datomerki8a5yj · Accepted Answer

Step 1The first question is much easier than you&#039;re making it. If f is differentiable, it has to be continuous. If f is continuous, then   f  (  0  )  =      lim          x      →      0        f  (  x  ).Step 2For the second question: note that      lim          h      →      0                  f      (      x      +      h      )      −      f      (      x      )        h    =      lim          h      →      0            1    h        (    f    (    x    +    h    )    −    f    (    x    )    )    =      lim          h      →      0            1    h        (                  f        (        x        )        +        f        (        h        )                    1        −        f        (        x        )        f        (        h        )              −    f    (    x    )    )

Antiderivative Theory Problem A function f is differentiable over its domain and has the following

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New Questions in Calculus 1