Finding the critical points of a function and the interval where it increases and decreases.I am having trouble finding the critical points of f ( x ) = ( x + 1 ) / x − 3 I found the derivative to be f ′ ( x ) = − 4 / ( x − 3 ) 2 .My next step was to equate my derivative to zero, but that does not seem to work as my x cancels out. Usually I would take the x-value(worked out by equating the derivative with zero) and substitute it into the original equation to get a y-value. This would then be the critical points. Is there anyone who could maybe help me out (maybe with an example or so) as I also have to find the intervals where the function is increasing and decreasing?

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Finding the critical points of a function and the interval where it increases and decreases.I am having trouble finding the critical points of  f  (  x  )  =  (  x  +  1  )      /        x    −    3  I found the derivative to be       f    ′    (  x  )  =  −  4      /    (  x  −  3      )    2  .My next step was to equate my derivative to zero, but that does not seem to work as my x cancels out. Usually I would take the x-value(worked out by equating the derivative with zero) and substitute it into the original equation to get a y-value. This would then be the critical points. Is there anyone who could maybe help me out (maybe with an example or so) as I also have to find the intervals where the function is increasing and decreasing?

nutnhonyl8 · Accepted Answer

Step 1Judging by the derivative you calculated, it appears the function is supposed to be   f  (  x  )  =            x      +      1              x      −      3      .Since the implicit domain of a rational function is the set of all real numbers except those that make the denominator equal to zero, the implicit domain of f is       Dom    f    =  {  x  ∈      R    ∣  x  ≠  3  }  =  (  −  ∞  ,  3  )  ∪  (  3  ,  ∞  ).The derivative of f is       f    ′    (  x  )  =  −      4          (      x      −      3              )        2              &amp;lt;  0 for every x in its domain, which tells us that the function is decreasing on the intervals   (  −  ∞  ,  3  ) and   (  3  ,  ∞  )Note that   f  (  x  )  =            x      +      1              x      −      3        =  1  +      4          x      −      3       so                               lim                      x            →            −            ∞                          f        (        x        )                            =                  lim                      x            →            −            ∞                                    (          1          +                      4                          x              −              3                                )                =        1                                      lim                      x            →                          3              +                                      f        (        x        )                            =                  lim                      x            →                          3              +                                                (          1          +                      4                          x              −              3                                )                =        −        ∞                                      lim                      x            →                          3              +                                      f        (        x        )                            =                  lim                      x            →                          3              +                                                (          1          +                      4                          x              −              3                                )                =        ∞                                      lim                      x            →            ∞                          f        (        x        )                            =                  lim                      x            →            ∞                                    (          1          +                      4                          x              −              3                                )                =        1            Step 2Thus, when   x  ∈  (  −  ∞  ,  3  ),   f  (  x  )  ∈  (  −  ∞  ,  1  ), and when   x  ∈  (  3  ,  ∞  ),   f  (  x  )  ∈  (  1  ,  ∞  ). Since the function assumes larger values in the interval   (  3  ,  ∞  ) than it does in the interval   (  −  ∞  ,  3  ) than it does in the interval   (  −  ∞  ,  3  ), it does not decrease over the union of the two intervals in which it is decreasing.If you define a critical point of a function f to be a point where       f    ′    (  x  )  =  0, then the function   f  :  (  −  ∞  ,  3  )  ∪  (  3  ,  ∞  )  →      R   defined by  f  (  x  )  =            x      +      1              x      −      3       does not have a critical point since       f    ′    (  x  )  &amp;lt;  0 for every x in its domain.If you use the alternative definition that a critical point of a function f is a point in its domain where       f    ′    (  x  )  =  0 or f′(x) does not exist, then the function   f  :  (  −  ∞  ,  3  )  ∪  (  3  ,  ∞  )  →      R   defined by   f  (  x  )  =            x      +      1              x      −      3       still does not have a critical point since the derivative is defined at every point of its domain.

Find the critical points of f(x)=((x+1))/(x-3).

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