Use the first and second derivatives of the function y = x^3-3x^2-9x+15 to find the intervals of increase and decrease. Help!?

1 Answer
Roy Ø.
Jun 3, 2018

The function is decreasing for #x in [-1, 3]#, and is decreasing for all x outside this interval, #x !in [-1, 3]#.

Explanation:

You may benefit from looking at some videos where the derivatives are explained, for instance

It is always a good idea to start with drawing a graph of your 3rd degree function (I've used Geogebra):
enter image source here
The graph shows that for #x in [-1, 3]# the function is decreasing, and it is decreasing for all x outside this interval, #x !in [-1, 3]#.

Let's use the 1st and 2nd derivatives to find this, as requested in the task:
#y=x^3-3x^2-9x+15#
#y'=3x^2-6x-9=3(x+1)(x-3)#
#y''=6x-6=6(x-1)#

#y'# gives the slope in each point of the curve, and #y''# tels if it's increasing or decreasing in the point. Therefore, for what values of x is #y'=0#?

The first derivative#y'=3(x+1)(x-3)#
We see that #y'=0# when #x=-1# and #x=3#, so the slope changes direction in these points.

Next: the 2nd derivative #y''=6(x-1)#.
#x=-1#: #y''=6(x-1)=6(-2)=-12<0#
A negative value tells us that the slope is changing from an increase to decrease

#x=3#: #y''=6(x-1)=6*2=12>0#
A positive value tells us that the slope is changing from a decrease to increase.

This is actually in agreement with what we could see from the graph.

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