Find min/max along with concavity, inflection points, and asymptotesFor the function ƒ ( x ) = x 2 ( x − 2 ) 2 I know the derivative is f ′ ( x ) = − 4 x ( x − 2 ) 3 and f ″ ( x ) = 8 ( x + 1 ) ( x − 2 ) 4 .Critical point is x = − 1 which is also the min.I think possible inflection point are x = 1 and x = 0.I need to find: 1)Find the vertical and horizontal asymptotes of ƒ(x).2) Find the intervals on which ƒ(x)‍ is increasing or decreasing.3) Find the critical numbers and specify where local maxima and minima of ƒ(x) occur.4) Find the intervals of concavity and the inflection points of ƒ(x).

Question

Find min/max along with concavity, inflection points, and asymptotesFor the function       ƒ    (  x  )  =            x      2              (      x      −      2              )        2            I know the derivative is       f    ′    (  x  )  =  −            4      x                      (        x        −        2        )            3       and       f    ″    (  x  )  =            8              (        x        +        1        )                            (        x        −        2        )            4      .Critical point is   x  =  −  1 which is also the min.I think possible inflection point are   x  =  1 and   x  =  0.I need to find: 1)Find the vertical and horizontal asymptotes of ƒ(x).2) Find the intervals on which ƒ(x)‍ is increasing or decreasing.3) Find the critical numbers and specify where local maxima and minima of ƒ(x) occur.4) Find the intervals of concavity and the inflection points of ƒ(x).

taguetzbo · Accepted Answer

Explanation:We have       f    ″    (  x  )  =  8                      x        +        1                              (          x          −          2          )                          4                     and       f    ″    (  x  )  =  0 if   x  =  −  1 and       f    ‴    (  x  )  =  −  24                      x        +        2                              (          x          −          2          )                          5                     plugging   x  =  −  1 in f′′′(x) we get       f    ‴    (  −  1  )  =      8    81    ≠  0 thus we have the inflection point   −  1  ,      1    9     x  =  0 gives no inflection point. We get the point (0,0) as a local minimum. Since   f  (  x  )  =  1  +      4          x      −      2        +      4          (      x      −      2              )        2             we get a horizontal asymptote   y  =  1. Since for   x  =  2 is the denominator equal to zero we have a vertical asymptote for   x  =  2 we get for   −  ∞  &amp;lt;  x  &amp;lt;  0 and   2  &amp;lt;  x  &amp;lt;  ∞ is the given function monotonnously decreasing and for   0  &amp;lt;  x  &amp;lt;  2 monotonously increasing.