What is the domain and range of #h(x)=(x-1)/(x^3-9x) #?

1 Answer
Jul 12, 2018

Answer:

Domain: # x in (-oo,-3)uu(-3,0)uu(0,3)uu(3,oo)#
Range : # h(x) in RR or (-oo,oo) #

Explanation:

#h(x)=(x-1)/(x^3-9 x) or h(x)=(x-1)/(x(x^2-9)# or

#h (x)=(x-1)/(x(x+3)(x-3)#

Domain: Possible input value of #x# , if denominator is

zero , the function is undefined .

Domain: #x# is any real value except #x=0 , x =-3 and x=3# .

In interval notation:

# x in (-oo,-3)uu(-3,0)uu(0,3)uu(3,oo)#

Range: Possible output of #h(x) # .When #x=1 ; h(x)=0#

Range : Any real value of #h(x) :. h(x) in RR or (-oo,oo) #

graph{(x-1)/(x^3-9x) [-10, 10, -5, 5]}[Ans]