Consider the graph of the function #f(x) = 2(x+3)^2+2#. Over which interval is the graph decreasing?

Question

6596 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-25T02:17:22

#f(x)# is decreasing over #(-oo, -3)#

#f(x)=2(x+3)^2+2#

First let's expand #f(x)#

#f(x) =2(x^2+6x+9)+2#

#= 2x^x+12x+20#

Note that #f(x)# is defined #forall x in RR#
#:.#the domain of #f(x)# is #(-oo, +oo)#

Also note that as #f(x)# is a parabola it will have a single turning point.

#f(x)# will be decreasing where #f'(x)<0#

#f'(x)= 4x+12# (Power rule)

Hence #f(x)# will be decreasing where: #4x+12<0#

i.e. where #x+3<0 -> x<-3#

Since the domain of #f(x)# is #(-oo, +oo)#, #f(x)# will be decreasing over: #(-oo, -3)#

This can be seen by the graph of #f(x)# below.

graph{2(x+3)^2+2 [-17.09, 11.38, -2.16, 12.08]}