Why Does f ( x ) = x x + 3 Only Have One Critical Point?I am trying to find the critical points of the function f ( x ) = x x + 3 , then by using the First Derivative Test, determine which ones are a local maximum, local minimum, or neither.Using the product rule, we get 3 ( x + 2 ) 2 x + 3 . So x = − 2 is a critical point. But isn't x = − 3 also a critical point because if x = − 3 then the derivative of f ( x ) wouldn't exist.To give an example of why I am thinking this, look at g ( x ) = x 2 / 3 ( 6 − x ) 1 / 3 .The derivative of g ( x ) is 4 − x x 1 / 3 ( 6 − x ) 2 / 3 , where x cannot equal 0 or 6.The critical points for g ( x ) are 0, 6, and 4.So why aren't − 2 and − 3 the critical points for f ( x )?

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Why Does   f  (  x  )  =  x      x    +    3   Only Have One Critical Point?I am trying to find the critical points of the function   f  (  x  )  =  x      x    +    3  , then by using the First Derivative Test, determine which ones are a local maximum, local minimum, or neither.Using the product rule, we get             3      (      x      +      2      )              2              x        +        3            . So   x  =  −  2 is a critical point. But isn&#039;t   x  =  −  3 also a critical point because if   x  =  −  3 then the derivative of   f  (  x  ) wouldn&#039;t exist.To give an example of why I am thinking this, look at   g  (  x  )  =      x          2              /            3        (  6  −  x      )          1              /            3      .The derivative of   g  (  x  ) is             4      −      x                      x                  1                      /                    3                    (      6      −      x              )                  2                      /                    3                    , where   x cannot equal   0 or   6.The critical points for   g  (  x  ) are   0,  6, and   4.So why aren&#039;t   −  2 and   −  3 the critical points for   f  (  x  )?

aushilfem8sif · Accepted Answer

In the hope of clarifying some of the confusion, the critical points of a differentiable function is any value in its domain where the derivative is    0 or undefined. Given the domain of   f  (  x  ) is   [  −  3  ,  ∞  ) and its derivative is:      f    ′    (  x  )  =                    3        (        x        +        2        )                    2                  x          +          3                    The critical points are:  −  2 because   −  2  ∈  [  −  3  ,  ∞  ) and       f    ′    (  −  2  )  =  0  −  3 because   −  3  ∈  [  −  3  ,  ∞  ) and       f    ′    (  −  3  ) is undefined.

sg101cp6vv · Accepted Answer

In order for   a to be a critical point (at least, under the definition I&#039;m familiar with),   a has to be in the interior of the domain of the function (that is, the function has to be defined on either side of   a); for any   x  &amp;lt;  −  3,   f  (  x  ) isn&#039;t defined (exercise: why?), so   −  3 can&#039;t be a critical point.

Why Does f ( x ) = x <msqrt> x + 3 </msqrt> Only Have One

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