Intervals of a derivativeI'm trying to find the intervals of the derivative of a function that I found on my book but I'm having some troubles understanding it, so I thought of trying to find some help here.The function is f ( x ) = arcsin ⁡ ( | x 2 + 3 x + 3 | − 1 ) .So I need to find the derivative of this function, study it and find the intervals (where it's increasing and where it's decreasing)This is the solution: − ( 2 x + 3 ) 1 − ( x 2 + 3 x + 3 ) 2 ( − 2 , − 1 ) and ( 2 x + 3 ) 1 − ( x 2 + 3 x + 1 ) 2 ( − 3 , − 2 ) ( − 1 , 0 ).So the function is decreasing in [ − 3 , − 2 ] and [ − 3 2 , − 1 ] while increasing in [ − 2 , − 3 2 ] and [ − 1 , 0 ].Here is where I am having some troubles:- How do I find those intervals? I know that I need to solve the equations, but I 'm stuck there- How can I find where the function is decreasing and increasing?

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Intervals of a derivativeI&#039;m trying to find the intervals of the derivative of a function that I found on my book but I&#039;m having some troubles understanding it, so I thought of trying to find some help here.The function is       f    (    x    )    =    arcsin    ⁡    (          |              x      2        +    3    x    +    3          |        −    1    )  .So I need to find the derivative of this function, study it and find the intervals (where it&#039;s increasing and where it&#039;s decreasing)This is the solution:   −            (      2      x      +      3      )              1      −      (              x        2            +      3      x      +      3              )        2                (  −  2  ,  −  1  ) and             (      2      x      +      3      )              1      −      (              x        2            +      3      x      +      1              )        2                 (  −  3  ,  −  2  )  (  −  1  ,  0  ).So the function is decreasing in   [  −  3  ,  −  2  ] and   [  −      3    2    ,  −  1  ] while increasing in   [  −  2  ,  −      3    2    ] and   [  −  1  ,  0  ].Here is where I am having some troubles:- How do I find those intervals? I know that I need to solve the equations, but I &#039;m stuck there- How can I find where the function is decreasing and increasing?

Makai Lang · Accepted Answer

Step 1First, note that the denominator is ALWAYS positive when it exists, because it&#039;s inside a square root.So now if you look at the numerator, for   −  (  2  x  +  3  ), for   (  −  2  ,  −  1  ), we see that it&#039;s positive when   (  2  x  +  3  )  &amp;lt;  0, so that means f(x) is increasing for   x  &amp;lt;      −          3      2      , or       (    −    2    ,          −              3        2              )  . The function is decreasing when   (  2  x  +  3  )  &amp;gt;  0 and the numerator is negative, which corresponds to the interval       (                  3        2            ,      −      1        )  .Step 2Now, we can look at the second piece. The numerator is positive, and the function is increasing for   x  &amp;gt;      −          3      2       and decreasing for   x  &amp;lt;      −          3      2      . Since we know that this function is only defined on (-1, 3), this means that f(x) is also increasing on (-1, 0) and decreasing on (-3, 2).The important thing was to consider each piece by itself, so now your answer is putting those 4 intervals together.

opositor5t · Accepted Answer

Step 1First of all, since       x    2    +  3  x  +  3  &amp;gt;  0  ,  x  ∈      R  , the expression of f is simply   f  (  x  )  =  arcsin  ⁡  (      x    2    +  3  x  +  2  ). Now, the domain of f is given by the points that satisfy   −  1  ≤      x    2    +  3  x  +  2  ≤  1. the left inequality is always satisfied, the right inequality is satisfied when       x    2    +  3  x  +  1  ≤  0, i.e.  x  ∈      [          1      2        (    −    3    −          5        )    ,          1      2        (    −    3    +          5        )    ]  Step 2Obviously, it does not make sense to analyse the sign of the derivative&#039;s expression when we are outside the domain of f. Finally, since       f    ′    (  x  )  =            2      x      +      3              1      −      (              x        2            +      3      x      +      2              )        2             you see that the derivative is negative for   x  &amp;lt;  −      3    2   and positive for   x  &amp;gt;  −      3    2  , but always remaining in the domain!

Intervals of a derivative. The function is f(x)=arcsin(|x^2+3x+3|-1)

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