Question #60236

2 Answers
Tony B
Jul 5, 2017

#-2 < x < 2#

Explanation:

Multiply two negatives expressions together and you get a positive.

#g(x)# is negative so #(x^2-4)# must also be negative.

Thus we need to solve for #x^2-4<0#

add 4 to both sides

#x^2<4#

square root both sides

#x<+-2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note that if #x=-2#, then we have #f(x)=((-2)^2-4)g(x)=0#

This does not satisfy the condition #f(x)>0# so #x=-2# is an excluded value. This is also true of and #x<-2# as it would make #f(x)# negative. Thus we have the restriction: #-2 < x < 2#

Name
Jul 5, 2017

#f(x)>0# for #-2 < x < 2#

Explanation:

The required range of values of #x# satisfies:
#f(x)>0#
or #(x^2-4).g(x)>0#
If #g(x)# is always negative, the the product #(x^2-4).g(x)# will only be positive (or >0) when #(x^2-4)# is also negative, i.e. when;
#(x^2-4)<0#
#x^2<4#
Taking the square root on both sides,
#absx<2#
or #-2 < x < 2#
This can be seen in the graph of #(x^2-4)#. The function takes a negative value only in the range #-2 < x < 2#
graph{x^2-4 [-5, 5, -2.5, 2.5]}
Therfore,
#f(x)>0# for #-2 < x < 2#

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