Some doubts with the sign of a derivativeGood evening to everyone. The derivative is defined in the following order: d d x f ( x ) = − x 2 − x + 11 ( x + 3 ) 2 e 2 − x for x &lt; − 3 d d x f ( x ) = x 2 + x − 11 ( x + 3 ) 2 e 2 − x for − 3 &lt; x &lt; 2 and d d x f ( x ) = x 2 + x + 1 ( x + 3 ) 2 e x − 2 for x &gt; 2. If you'll do the sign of the first one it'll be x 2 + x − 11 &lt; 0 therefore only x 1 = − 1 + 45 2 verifies the conditions, for the second one it is x 2 + x − 11 &gt; 0 therefore x 1 = − 1 + 45 2 and x 1 = − 1 − 45 2 , the last one x 2 + x + 1 &gt; 0 has the solutions x 1 = − 1 − i 3 2 and x 2 = − 1 + i 3 2 (here I think that I did a mistake). Therefore our function should increase on the interval ( − ∞ , − 1 − 45 2 ) then decrease on the interval ( − 1 − 45 2 , − 1 + 45 2 ) then increase again on the interval ( − 1 + 45 2 , − 1 − i 3 2 ) and decrease again on ( − 1 + i 3 2 , ∞ ). But Instead the solutions of the third equation are x 1 = − 1 − 5 2 and x 2 = − 1 + 5 2 . And it is decreasing on ( − ∞ , − 1 − 45 2 ), then increasing on ( − 1 − 45 2 , − 3 ) then decreasing on (-3,2) then increasing again on ( 2 , ∞ ). I don't get where I'm wrong.. neither with the results of quadratic equation nor with the intervals of monotony(sign) of the derivative.

Question

Some doubts with the sign of a derivativeGood evening to everyone. The derivative is defined in the following order:      d          d      x        f  (  x  )  =            −              x        2            −      x      +      11                      (        x        +        3        )            2            e          2      −      x         for   x  &amp;lt;  −  3        d          d      x        f  (  x  )  =                    x        2            +      x      −      11                      (        x        +        3        )            2            e          2      −      x       for   −  3  &amp;lt;  x  &amp;lt;  2 and       d          d      x        f  (  x  )  =                    x        2            +      x      +      1                      (        x        +        3        )            2            e          x      −      2       for   x  &amp;gt;  2. If you&#039;ll do the sign of the first one it&#039;ll be             x      2        +    x    −    11    &amp;lt;  0 therefore only       x    1    =            −      1      +              45              2   verifies the conditions, for the second one it is             x      2        +    x    −    11    &amp;gt;  0 therefore       x    1    =            −      1      +              45              2   and       x    1    =            −      1      −              45              2  , the last one             x      2        +    x    +    1    &amp;gt;  0 has the solutions       x    1    =            −      1      −      i              3              2   and       x    2    =            −      1      +      i              3              2   (here I think that I did a mistake). Therefore our function should increase on the interval   (  −  ∞  ,            −      1      −              45              2    ) then decrease on the interval   (            −      1      −              45              2    ,            −      1      +              45              2    ) then increase again on the interval   (            −      1      +              45              2    ,            −      1      −      i              3              2    ) and decrease again on   (            −      1      +      i              3              2    ,  ∞  ). But Instead the solutions of the third equation are       x    1    =            −      1      −              5              2   and       x    2    =            −      1      +              5              2  . And it is decreasing on   (  −  ∞  ,            −      1      −              45              2    ), then increasing on   (            −      1      −              45              2    ,  −  3  ) then decreasing on (-3,2) then increasing again on   (  2  ,  ∞  ). I don&#039;t get where I&#039;m wrong.. neither with the results of quadratic equation nor with the intervals of monotony(sign) of the derivative.

Kaeden Bishop · Accepted Answer

Step 1We have a root for the first derivative at around   x  ≈  −  3.854 the other is negligible as the derivative is not defined with the particular function at that point.With   x  &amp;lt;  −            1      +      3              5              2    ,      f    ′    (  x  )  &amp;lt;  0. Thus, f(x) is decreasing for   x  ∈  (  −  ∞  ,  −            1      +      3              5              2    ) and increasing for   x  ∈  (  −            1      +      3              5              2    ,  −  3  ). (I substituted a random value for   x  &amp;lt;  −            1      +      3              5              2   and found that       f    ′    (  x  )  &amp;lt;  0).Step 2For   −  3  &amp;lt;  x  &amp;lt;  2, the derivative defined as the same roots so now the root around   x  ≈  2.854 is relevant and the other is negligible.With   −  3  &amp;lt;  x  &amp;lt;  2  ,      f    ′    (  x  ) is negative. Thus, f(x) is decreasing from   x  ∈  (  −  3  ,  2  ).For   x  &amp;gt;  2 the derivative defined does not have any real roots. With a random check, we find that at any point   x  &amp;gt;  2  ,      f    ′    (  x  )  &amp;gt;  0. Thus, f(x) is always increasing on   x  ∈  (  2  ,  ∞  )

The derivative is defined in the following order: d/dxf(x)=(-x^2-x+11)/(x+3)^2 for x<-3

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