How do you find the domain and range of #f(x)= (3x-1)/(sqrt(x^2+x-2))#?

Question

Eli Assouline · Accepted Answer

Domain : #x : (-oo,-2) uu (1,oo)#
Range : #f(x) : (-3, -oo) uu (3,oo)# #f(x)= (3x-1)/sqrt(x^2+x-2) = (3x-1)/sqrt((x+2)(x-1))# Domain: denominator must not be #0# and under root must not be #<0# So #x+2!=0 :. x != -2# and #x-1!=0 :. x != 1 # For under root calculation: critical points are #x=-2 and x= 1# when #x <-2 , (x+2)* (x-1) = (-*- )= + # when # -2 < x < 1 , (x+2)* (x-1) = (+*- ) = - # when #x >1 , (x+2)* (x-1) = (=*+ )= + # So in domain : #x < -2 and x> 1 or (-oo,-2) uu (1,oo)# Horizontal asymptote is at #y= 3/(+-1)=+-3# So range : # f(x) : (-3, -oo) uu (3,oo)# graph{(3x-1)/(x^2+x-2)^0.5 [-20, 20, -10, 10]} [Ans]

Savannah Anderson · Answer

undefined To find the domain and range of the function $ f(x) = \frac{3x - 1}{\sqrt{x^2 + x - 2}} $:

1. **Domain**:
   - The function is defined for all real numbers except where the denominator becomes zero, as division by zero is undefined. So, we need to find the values of $ x $ that make the denominator zero.
   - The expression inside the square root, $ x^2 + x - 2 $, factors as $ (x + 2)(x - 1) $.
   - Therefore, the function is undefined when $ x = -2 $ and $ x = 1 $.
   - Thus, the domain of $ f(x) $ is all real numbers except $ x = -2 $ and $ x = 1 $, expressed in interval notation as $ (-\infty, -2) \cup (-2, 1) \cup (1, \infty) $.

2. **Range**:
   - To find the range, we need to consider the behavior of the function as $ x $ approaches positive and negative infinity.
   - As $ x $ approaches positive infinity, both the numerator and the denominator grow without bound. Since the degree of the numerator is less than the degree of the denominator, the function approaches zero.
   - As $ x $ approaches negative infinity, both the numerator and the denominator grow without bound, but in this case, the denominator becomes negative. Therefore, the function approaches negative infinity.
   - Hence, the range of $ f(x) $ is all real numbers, expressed in interval notation as $ (-\infty, \infty) $.