Find the interval in which f(x) increases and decreases.Let f ( x ) = 2 x 3 − 9 x 2 + 12 x + 6 so f ′ ( x ) = 6 x 2 − 18 x + 12 = 6 ( x − 1 ) ( x − 2 ).I need the intervals in which f(x) strictly increases, f ′ ( x ) &gt; 0 when x &lt; 1 and x &gt; 2 and thus f(x) strictly increases in these intervals and f ′ ( x ) &lt; 0 when 1 &lt; x &lt; 2 so f(x) strictly decreases in this interval.My Question:What about at points x = 1 , 2.If f ′ ( x ) = 0 at points (not intervals) then f(x) can still be considered strictly monotone. And it also seems reasonable (I'll add the reason below) to include the points 1,2 in the intervals of increase (IOI, for short) and decreases (IOD).Eg: Take x = 1. Let's say I include this point in both IOI and IOD. So IOI is now, x ∈ ( − ∞ , 1 ] and you can see that it doesn't contradict the definition of "strictly increasing function in interval" either. Take any p , q ∈ ( − ∞ , 1 ] , p &gt; q ⇒ f ( p ) &gt; f ( q ) similarly my IOD, now would be x ∈ [ 1 , 2 ] (notice I included 2) and it still follows the definition.As I understand that, definition of monotonicity functions at a point, would now get in the way.I can't include x = 1 because there exists no h &gt; 0 such that taking p , q ∈ ( 1 − h , 1 + h ) ⇏ f ( p ) &gt; f ( q ) similarly, I can't add x = 1 in interval of decrease either.If I'd have to pick, my intuition would lead me to pick the second one, but I can't see why I should reject the first one either since it actually follows the definition of strictly increasing function in interval.

Question

Find the interval in which f(x) increases and decreases.Let   f  (  x  )  =  2      x    3    −  9      x    2    +  12  x  +  6 so       f    ′    (  x  )  =  6      x    2    −  18  x  +  12  =  6  (  x  −  1  )  (  x  −  2  ).I need the intervals in which f(x) strictly increases,       f    ′    (  x  )  &amp;gt;  0 when   x  &amp;lt;  1 and   x  &amp;gt;  2 and thus f(x) strictly increases in these intervals and       f    ′    (  x  )  &amp;lt;  0 when   1  &amp;lt;  x  &amp;lt;  2 so f(x) strictly decreases in this interval.My Question:What about at points   x  =  1  ,  2.If       f    ′    (  x  )  =  0 at points (not intervals) then f(x) can still be considered strictly monotone. And it also seems reasonable (I&#039;ll add the reason below) to include the points 1,2 in the intervals of increase (IOI, for short) and decreases (IOD).Eg: Take   x  =  1. Let&#039;s say I include this point in both IOI and IOD. So IOI is now,   x  ∈  (  −      ∞    ,  1  ] and you can see that it doesn&#039;t contradict the definition of &quot;strictly increasing function in interval&quot; either. Take any   p  ,  q  ∈  (  −      ∞    ,  1  ]  ,  p  &amp;gt;  q  ⇒  f  (  p  )  &amp;gt;  f  (  q  ) similarly my IOD, now would be   x  ∈  [  1  ,  2  ] (notice I included 2) and it still follows the definition.As I understand that, definition of monotonicity functions at a point, would now get in the way.I can&#039;t include   x  =  1 because there exists no   h  &amp;gt;  0 such that taking   p  ,  q  ∈  (  1  −  h  ,  1  +  h  )  ⇏  f  (  p  )  &amp;gt;  f  (  q  ) similarly, I can&#039;t add   x  =  1 in interval of decrease either.If I&#039;d have to pick, my intuition would lead me to pick the second one, but I can&#039;t see why I should reject the first one either since it actually follows the definition of strictly increasing function in interval.

lufi8c · Accepted Answer

Step 1Let J be any (open, halfopen, closed / bounded, unbounded) interval and   g  :  J  →      R   be a function. Then g is strictly increasing on J if for all   x  ,  y  ∈  J such that   x  &amp;lt;  y we have   g  (  x  )  &amp;lt;  g  (  y  ). You may also define that g is strictly increasing in a point   p  ∈  J if there is   ε  &amp;gt;  0 such that g is strictly increasing on   J  ∩  (  p  −  ε  ,  p  +  ε  ).You have a function   f  :      R    →      R   and ask for the (maximal) intervals J on which f strictly increases or decreases.Step 2If you interpret this in the sense that f∣J has this property, then you get the result which you call reasonable. I share this point of view.However, you could also interpret the question in the sense that f should strictly increase or decrease in all points of J which gives your &quot;but&quot;.There is no contradiction, but only different interpretations, and you have to decide which you prefer.

Let f(x)=2x^3-9x^2+12x+6 so f′(x)=6x^2-18x+12=6(x-1)(x-2). I need the intervals in which f(x) strictly increases, f′(x)>0 when x<1 and x>2 and thus f(x) strictly increases in these intervals and f′(x)<0 when 1<x<2 so f(x) strictly decreases in this interval. What about at points x=1,2.

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