How do you find the domain and range of #1 /( x^3-9x)#?

Question

Hazel Anglin · Accepted Answer

#x inRR,x!=0,x!=+-3#
#y inRR,y!=0# #"the denominator of " y=1/(x^3-9x)" cannot be zero"# #"as this would make y undefined"# #"equating the denominator to zero and solving gives the"# #"values that x cannot be"# #"solve " x^3-9x=0rArrx(x-3)(x+3)=0# #rArrx=0,x=+-3larrcolor(red)"excluded values"# #"domain is " x inRR,x!=0,x!=+-3# #"to find any excluded values in the range"# #"consider the horizontal asymptote of the function"# #"divide terms on numerator/denominator by the highest"# #"power of x, that is " x^3# #y=(1/x^3)/(x^3/x^3-(9x)/x^3)=(1/x^3)/(1-9/x^2)# as #xto+-oo,yto0/(1-0)=0larrcolor(red)" excluded value"# #"range is " y inRR,y!=0#

Ethan Adkins · Answer

undefined To find the domain of the function, set the denominator $ x^3 - 9x $ not equal to zero and solve for $ x $. This gives $ x \neq 0 $ and $ x \neq 3 $ as restrictions on the domain.

For the range, consider the behavior of $ \frac{1}{x^3 - 9x} $ as $ x $ approaches positive or negative infinity. As $ x $ approaches positive or negative infinity, $ x^3 $ dominates $ 9x $, so $ x^3 - 9x $ approaches positive or negative infinity, respectively. Therefore, the range is all real numbers except for $ 0 $.