Let f(x)=(2x−1)x and g(x)=(x−1)x. Then, the domain of the function h(x)=f(g(x))+g(f(x)) is all real numbers except
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Official Correct Answer: B. First, the domain of a function includes all the real numbers for which the function is defined. Since we're dealing with rational functions here, the main concern is division by zero. So, We find all the x-values where the denominators in f(g(x)) and g(f(x)) become zero because those points will be excluded from the domain.
Let me start by understanding what h(x) is. It's the sum of two composite functions: f(g(x)) and g(f(x)). So, I need to compute both f(g(x)) and g(f(x)) separately and then add them together.
Starting with f(g(x)). Since f(x) = x/(2x - 1), replacing x with g(x) gives f(g(x)) = g(x)/(2g(x) - 1). Similarly, g(f(x)) would be f(x)/(f(x) - 1). So, I need to compute both of these expressions.
Compute f(g(x)) first. Given that g(x) = x/(x - 1), substituting this into f(g(x)) gives:
f(g(x)) = [x/(x - 1)] / [2*(x/(x - 1)) - 1]
Simplify the denominator:
2*(x/(x - 1)) - 1 = (2x)/(x - 1) - 1 = (2x - (x - 1))/(x - 1) = (2x - x + 1)/(x - 1) = (x + 1)/(x - 1)
So, f(g(x)) becomes [x/(x - 1)] / [(x + 1)/(x - 1)] = [x/(x - 1)] * [(x - 1)/(x + 1)] = x/(x + 1)
Wait, that simplifies nicely! The (x - 1) terms cancel out, leaving x/(x + 1). So, f(g(x)) is x/(x + 1). That seems straightforward.
Now, moving on to g(f(x)). Since g(x) = x/(x - 1), replacing x with f(x) gives:
g(f(x)) = f(x)/(f(x) - 1) = [x/(2x - 1)] / [x/(2x - 1) - 1]
Simplify the denominator:
x/(2x - 1) - 1 = [x - (2x - 1)]/(2x - 1) = (x - 2x + 1)/(2x - 1) = (-x + 1)/(2x - 1)
So, g(f(x)) becomes [x/(2x - 1)] / [(-x + 1)/(2x - 1)] = [x/(2x - 1)] * [(2x - 1)/(-x + 1)] = x/(-x + 1) = -x/(x - 1)
Hmm, interesting. So, g(f(x)) simplifies to -x/(x - 1). That's a bit simpler than I expected.
Now, h(x) is the sum of f(g(x)) and g(f(x)), which is:
h(x) = x/(x + 1) + (-x)/(x - 1) = x/(x + 1) - x/(x - 1)
To combine these two fractions, I need a common denominator. The denominators are (x + 1) and (x - 1), so the common denominator is (x + 1)(x - 1). Let's rewrite each term:
x/(x + 1) = x(x - 1)/[(x + 1)(x - 1)] = (x^2 - x)/[(x + 1)(x - 1)]
Similarly, -x/(x - 1) = -x(x + 1)/[(x + 1)(x - 1)] = (-x^2 - x)/[(x + 1)(x - 1)]
Adding these together:
(x^2 - x) + (-x^2 - x) = x^2 - x - x^2 - x = (-2x)
So, h(x) simplifies to (-2x)/[(x + 1)(x - 1)] = (-2x)/(x^2 - 1)
Wait, so h(x) is (-2x)/(x^2 - 1). That seems much simpler. Now, to find the domain of h(x), we need to find all real numbers x where h(x) is defined. Since h(x) is a rational function, it's defined everywhere except where the denominator is zero.
So, set the denominator equal to zero:
x^2 - 1 = 0 => x^2 = 1 => x = 1 or x = -1
Therefore, x cannot be 1 or -1 because the denominator becomes zero, making h(x) undefined at these points.
But wait, I also need to consider the domains of f(g(x)) and g(f(x)) individually because h(x) is the sum of these two functions. Even if h(x) simplifies to a function defined at x = 1 or x = -1, if either f(g(x)) or g(f(x)) is undefined at those points, then h(x) is also undefined there.
So, let's check the domains of f(g(x)) and g(f(x)) separately.
Starting with f(g(x)) = x/(x + 1). The denominator is x + 1, which is zero when x = -1. So, x cannot be -1.
Next, g(f(x)) = -x/(x - 1). The denominator is x - 1, which is zero when x = 1. So, x cannot be 1.
Additionally, when we computed f(g(x)) and g(f(x)), we had to ensure that the intermediate steps didn't involve division by zero. Let's verify that.
For f(g(x)):
We had g(x) = x/(x - 1). So, the denominator x - 1 cannot be zero, which means x ≠ 1. Also, in f(g(x)), after substituting, we had 2g(x) - 1 in the denominator. Let's check if that can be zero:
2g(x) - 1 = 0 => 2*(x/(x - 1)) - 1 = 0 => (2x)/(x - 1) -