+0

# domain

+1
4
2

If $$f(x)$$ is a function whose domain is [-8,8], and $$g(x)=f\left((x^2 - 2)/(x + 1)\right)$$, then the domain of $$g(x)$$ is an interval of what width?

Oct 18, 2023
edited by Alan  Oct 19, 2023

#1
-2

The domain of g(x) is all real numbers x such that the expression x+1x2−2​ is defined and evaluates to a value in the domain of f(x).

The expression x+1x2−2​ is undefined when x=−1. Also, the expression x+1x2−2​ is defined and evaluates to 0 when x=−2. Therefore, the intersection of the domain of x+1x2−2​ with the domain of f(x) is the interval (−∞,−2)∪(−1,∞). However, the expression f((x2−2)/(x+1)) is also undefined when x=1. Therefore, the domain of g(x) is the interval (−∞,−2)∪(−1,1)∪(1,∞).

The width of this interval is 18−(−8)+1+1=18​.

Oct 18, 2023
#2
+2

Look at the function of g(x). How will changing f(x) to f((x^2 - 2)/(x + 1)) affect the domain?

The numerator of (x^2 - 2) only affects the range since all real numbers x can be plugged into it. From this, the domain is unaffected and stays at [-8,8].

Now, we must consider the denominator of (x + 1). Remember that the domain of a function is the set of possible values for x in the function. What is not a possible value here? Well, dividing by 0 would result in an undefined y value, so (x + 1) ≠ 0. Solving for x yields -1. This is the only x value that is not valid since all other real numbers x can be plugged into the denominator. Thus, the denominator constricts our new domain to [-8,1) U (1,8].

Now that we have the domain of f((x^2 - 2)/(x + 1)) which is equal to g(x), we can solve for the width of the interval of the domain of g(x). [-8 to 1) has a width of 9, and (1, 8] has a width of 7.  9+7=16, so 16 is our answer.

Oct 18, 2023