1. To find the inverse of g(x)=1/3x-7, we can start by replacing g(x) with y:
y = 1/3x - 7
Next, we can swap x and y and solve for y:
x = 1/3y - 7
To isolate y, we can add 7 to both sides:
x + 7 = 1/3y
Multiplying both sides by 3 gives:
3x + 21 = y
So, the inverse function of g(x)=1/3x-7 is h(x) = 3x + 21.
The domain of g(x) is all real numbers except x = 0 since dividing by zero is undefined. Therefore, the domain of the inverse function h(x) is also all real numbers except x = 0.
To find the range of g(x), we can observe that as x approaches infinity or negative infinity, g(x) approaches positive infinity or negative infinity, respectively. Therefore, the range of g(x) is all real numbers except for zero. Since the inverse function h(x) is the same function as g(x), its range will also be all real numbers except for zero.
2. To determine if f(x)=5x+1/x and g(x)=x/5x+1 are inverse functions, we need to check if f(g(x)) = x and g(f(x)) = x for all values in their respective domains.
First, let's find f(g(x)):
f(g(x)) = 5(g(x)) + 1/(g(x))
= 5(x/5x + 1) + 1/(x/5x + 1)
= x + 1/(x + 1)
We can simplify this expression as follows:
f(g(x)) = x + 1/(x + 1)
= (x(x + 1) + 1)/(x + 1)
= (x^2 + x + 1)/(x + 1)
f(g(x)) is not equal to x, so f(x)=5x+1/x and g(x)=x/5x+1 are not inverse functions.
We know they are not inverse functions because the composition of f(g(x)) does not simplify to just x.
1. Find the inverse of g(x)=1/3x-7 and then find its domain and range
2. Determine whether f(x)=5x+1/x and g(x)=x/5x+1 are inverse functions. Explain how you know.
1 answer