How do you check which intervals a cubic function will increase and in which intervals it will decrease?I was trying to find the intervals in which the cubic function 4 x 3 − 6 x 2 − 72 x + 30 would be strictly increasing and strictly decreasing.I managed to get the fact that at the values {-2,3} the differential of the function is zero. However this divides the function in three intervals, how can i know which intervals the function increases in and which intervals the function decreases in?Note: I know you could simply plot the function. I was hoping for a more analytical method.

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How do you check which intervals a cubic function will increase and in which intervals it will decrease?I was trying to find the intervals in which the cubic function   4      x    3    −  6      x    2    −  72  x  +  30 would be strictly increasing and strictly decreasing.I managed to get the fact that at the values {-2,3} the differential of the function is zero. However this divides the function in three intervals, how can i know which intervals the function increases in and which intervals the function decreases in?Note: I know you could simply plot the function. I was hoping for a more analytical method.

Leon Clark · Accepted Answer

Step 1Let&#039;s look at this in a little more depth. We have the function   f  (  x  )  =  4      x    3    −  6      x    2    −  72  x  +  30 with derivative       f    ′    (  x  )  =  12      x    2    −  12  x  −  72.Step 2We know the function is either going &quot;up-down-up&quot; or &quot;down-up-down&quot; (this you can say if you found two distict zeroes in the derivative- if not, then the cubic goes only &quot;up&quot; or only &quot;down&quot;), since you&#039;ve found       f    ′    (  −  2  )  =      f    ′    (  3  )  =  0. We can look at the leading coefficient to check where the function goes up and down (if the leading coefficient is positive, then it is &quot;up-down-up&quot; and otherwise, it is &quot;down-up-down&quot;. I might make this a little more intuitive if I say: in, for example,       x    3    −      x    2  , the       x    3   will always win of   −      x    2  , so in the end, the function will increase). We can also compute the derivative in a specific point in one of the intervals   (  −  ∞  ,  −  2  ),   (  −  2  ,  3  ) or   (  3  ,  ∞  ), for example, in   (  −  2  ,  3  ), to get       f    ′    (  0  )  =  −  72 to know that the function is decreasing on that interval, and so it must be &quot;up-down-up&quot;.

Ronin Tran · Accepted Answer

Step 1Whether a differentiable function is increasing or decreasing (or stationary) at a point is (essentially) determined by the sign of its derivative. For a cubic polynomial function   p  (  x  )  =  a      x    3    +  b      x    2    +  c  x  +  d  ,, the derivative is       p    ′    (  x  )  =  3  a      x    2    +  2  b  x  +  c  ..Step 2Thus, the character of the roots of p′, that is, the critical points of p, is determined completely by the discriminant   Δ  :=  (  (  2  b      )    2    )  −  4  (  3  a  )  (  c  )  =  4      b    2    −  12  a  c of p&#039;:- If   Δ  &amp;gt;  0, then p′ has two distinct roots,       r    −    &amp;lt;      r    +  . If   a  &amp;gt;  0 (in particular, if it is monic), then since       lim          x      →      ±      ∞            p    ′    (  x  )  =  +  ∞, we conclude that p is increasing where   x  &amp;lt;      r    −   and   x  &amp;gt;      r    +  , decreasing where       r    −    &amp;lt;  x  &amp;lt;      r    +  , and hence has a local maximum at       r    −   and a local minimum at       r    +  .- If   Δ  =  0, then p′ has a double root, r. If   a  &amp;gt;  0, we conclude that p is increasing where   x  ≠  r and that       p    ′    (  r  )  =  0, so that p has an inflection point at r, and has no minimum or maximum. (In fact, p is strictly increasing everywhere, as   x  &amp;lt;  y implies   p  (  x  )  &amp;lt;  p  (  y  ).)- If   Δ  &amp;lt;  0, then p′ has no roots. If   a  &amp;gt;  0,  p is increasing everywhere.(All of these statements are reversed appropriately when   a  &amp;lt;  0.)Example For our polynomial   p  (  x  )  =  4      x    3    −  6      x    2    −  72  x  +  30  ,, computing gives   Δ  =  3600  &amp;gt;  0, and we are in the first case.

Find the intervals in which the cubic function 4x^3-6x^2-72x+30 would be strictly increasing and strictly decreasing.

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