f(x)=(1+x)/(1-x),g(x)=x/(1-x), find the simplest form of f[g(x)] and state the domain.
My work:
(1+ (x/(1-x)))/(1-(x/(1-x)))
(1+x)/(1-x)/(1-x)/(1-x)
f (g(x) ) = (1 + g(x) )/(1 - g(x) )
= (1 + x/(1-x) / (1 - x/(1-x) )
multiply top and bottom by (1-x)
= (1 - x + x) / (1 - x - x)
= 1/(1 - 2x)
check:
let x = 2
g(2) = 2/(1-2) = -2
f(g(2)) = f(-2)
= (1-2)/(1+2) = -1/3
in mine f(g(2)) = 1/(1-4) = -1/3
it is "highly likely" that my answer is correct
Reiny's answer is correct, according to
(1 + x/(1-x) / (1 - x/(1-x) )
The domain is all reals except x = 1/2 or 1
Even though the domain of 1/(1-2x) only excludes x = 1/2, we also need to exclude x=1 since f(1) and g(1) are undefined, so f(g(1)) cannot be defined either.
To find the simplest form of f[g(x)], we need to substitute g(x) into f(x) and simplify.
Start by replacing x in f(x) with g(x):
f[g(x)] = f(x/(1-x))
Now, we substitute g(x) = x/(1-x) into f(x):
f[g(x)] = f(x/(1-x)) = (1 + (x/(1-x))) / (1 - (x/(1-x)))
To simplify this expression, we need to get a common denominator in both the numerator and the denominator:
f[g(x)] = (1(1-x) + x) / (1(1-x) - x)
Expanding the numerator and denominator gives us:
f[g(x)] = (1 - x + x) / (1 - x - x)
Simplifying further, we can combine like terms in both the numerator and denominator:
f[g(x)] = 1 / (1 - 2x)
Therefore, the simplest form of f[g(x)] is 1 / (1 - 2x).
Now, let's determine the domain of f[g(x)].
The domain is the set of all values of x for which the function is defined. In this case, we need to consider any values of x that would result in division by zero or make the expression in the denominator undefined.
In the expression 1 / (1 - 2x), the denominator would be equal to zero when:
1 - 2x = 0
Solving this equation for x gives us:
2x = 1
x = 1/2
Therefore, the domain of f[g(x)] is all real numbers except x = 1/2.