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. 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. To find the inverse of the function g(x) = (1/3)x - 7, we can follow these steps:
Step 1: Replace g(x) with y: y = (1/3)x - 7.
Step 2: Swap x and y: x = (1/3)y - 7.
Step 3: Solve for y: Multiply both sides by 3 to eliminate the fraction: 3x = y - 21. Then, add 21 to both sides: y = 3x + 21.
Therefore, the inverse of g(x) is g^(-1)(x) = 3x + 21.
Now, let's find the domain and range of g^(-1)(x):
The domain of g^(-1)(x) is the range of g(x), and the range of g^(-1)(x) is the domain of g(x).
For the function g(x) = (1/3)x - 7, its domain is all real numbers since there are no restrictions on x.
The range of g(x) is also all real numbers because the function is a linear function with a slope of 1/3, which means it covers all possible outputs.
Therefore, the domain of g^(-1)(x) is all real numbers, and the range of g^(-1)(x) is also all real numbers.
2. To determine if f(x) = 5x + 1/x and g(x) = x/(5x + 1) are inverse functions, we can check if their compositions equal to the identity function.
First, let's find the composition of f(g(x)):
f(g(x)) = 5(g(x)) + 1/(g(x)) = 5(x/(5x + 1)) + 1/(x/(5x + 1))
Simplifying further:
f(g(x)) = 5x/(5x + 1) + (5x + 1)/x
To combine the two terms, we need to have a common denominator:
f(g(x)) = (5x^2 + x + (5x + 1)(5x + 1))/(x(5x + 1))
Expanding and simplifying:
f(g(x)) = (5x^2 + x + 25x^2 + 10x + 1)/(5x^2 + x)
f(g(x)) = (30x^2 + 11x + 1)/(5x^2 + x)
Now, let's find the composition of g(f(x)):
g(f(x)) = (f(x))/(5f(x) + 1) = (5x + 1/x)/(5(5x + 1) + 1) = (5x + 1/x)/(25x + 6)
To combine the two terms, we need to have a common denominator:
g(f(x)) = (x(5x + 1) + 1)/(x(25x + 6))
Simplifying further:
g(f(x)) = (5x^2 + x + 1)/(25x^2 + 6x)
From the analysis above, we can see that f(g(x)) = (30x^2 + 11x + 1)/(5x^2 + x) and g(f(x)) = (5x^2 + x + 1)/(25x^2 + 6x).
Since f(g(x)) and g(f(x)) are not equal to x, they are not the identity function. Therefore, f(x) = 5x + 1/x and g(x) = x/(5x + 1) are not inverse functions.