Ok I got another one I can figure out
The one-to-one function f is defined by f(x)=(4x-1)/(x+7).
Find f^-1, the inverse of f. Then, give the domain and range of f^-1 using interval notation.
f^-1(x)=
Domain (f^-1)=
Range (f^-1)=
Any help is greatly appreciated.
f(x)=(4x-1)/(x+7)
y = (4x-1)/(x+7)
Rewrite as:
y = (4x)/(x+7)- 1/(x+7)
Multiply both sides by x+7:
(x + 7)y = 4x - 1
Expand out terms of the left hand side:
xy + 7y = 4x - 1
xy - 4x = -7y - 1
x(y - 4) = -7y - 1
Divide both sides by y - 4:
x = (-7y - 1)/(y - 4)
f^-1 = (-7x - 1)/(x - 4)
Can you do the domain and range now?
I don't know if this is right but this is what I came up with.
f^-1=(-7x+1)/(x-4)
domain f(^-1)=(-inf,-7)U(-7,inf)
range f(^-1)=(-inf,4)U(4,inf)
Let me know if it looks right. Thanks
To find the inverse of a function, we need to swap the roles of x and y and then solve for y. Let's follow these steps to find the inverse of f(x):
Step 1: Swap x and y in the equation:
x = (4y - 1) / (y + 7)
Step 2: Solve for y:
Multiply both sides by (y + 7) to eliminate the denominator:
x(y + 7) = 4y - 1
xy + 7x = 4y - 1
xy - 4y = -7x - 1
y(x - 4) = -7x - 1
y = (-7x - 1)/(x - 4)
So, the inverse of f(x) is f^(-1)(x) = (-7x - 1)/(x - 4).
Now, let's determine the domain and range of f^(-1) using interval notation:
Domain (f^-1): The domain of f^-1 refers to the values of x for which the inverse function is defined. In this case, f^(-1) is defined for all values of x except x = 4 (which causes division by zero in the denominator). Therefore, the domain of f^(-1) is (-∞, 4) U (4, ∞).
Range (f^-1): The range of f^-1 refers to the set of all possible output values of the inverse function. To determine the range, we need to consider the behavior of the function as x approaches positive or negative infinity. As x approaches positive infinity, (-7x - 1)/(x - 4) approaches -7, and as x approaches negative infinity, (-7x - 1)/(x - 4) approaches 7. So, the range of f^(-1) is (-∞, -7] U [7, ∞).
Therefore:
f^-1(x) = (-7x - 1)/(x - 4)
Domain (f^-1) = (-∞, 4) U (4, ∞)
Range (f^-1) = (-∞, -7] U [7, ∞)