Determine whether the following function is a one-to-one function,and if the function is a one-to-one, find a formula for the inverse.
F(x)=5/(x+3)
Inverse:
step 1. switch the x and y variables
y = 5/(x+3) -----> x = 5/(y+3)
step 2: solve this new equation for y
xy + 3x = 5
xy = 5-3x
y = (5-3x)/x
I was able to find the inverse, but both the original and the inverse have restrictions, so
it is not a one-to-one function
e.g. In the original , if x = -3, there is no value of y, so: not a 1-1
in the inverse, if x = 0 , there is no value of y, so: not a 1-1
Wolfram verification:
http://www.wolframalpha.com/input/?i=plot+y+%3D+5%2F%28x%2B3%29+%2C+x+%3D+5%2F%28y%2B3%29+%2C+y+%3D+x
To determine whether the function F(x) = 5/(x+3) is a one-to-one function, we need to check if every input value, x, has a unique output value, y.
To do this, we can use the horizontal line test. If every horizontal line intersects the graph of the function at most once, then the function is one-to-one.
Let's calculate the derivatives of F(x) to see if the function is one-to-one. The derivative of F(x) is given as:
F'(x) = -5/(x+3)^2
The derivative is never zero for all real values of x. Therefore, the function does not have any critical points, and it is always increasing or decreasing. This means that there are no horizontal lines that intersect the graph of the function more than once. Therefore, the function F(x) = 5/(x+3) is a one-to-one function.
To find the formula for the inverse function, we can follow these steps:
1. Replace F(x) with y: y = 5/(x+3)
2. Swap x and y: x = 5/(y+3)
3. Solve for y: Multiply both sides by (y+3) and rearrange the equation:
x(y+3) = 5
xy + 3x = 5
xy = 5 - 3x
y = (5 - 3x)/x
y = 5/x - 3
Therefore, the formula for the inverse function, denoted as F^(-1)(x), is:
F^(-1)(x) = 5/x - 3
To determine if the function F(x) = 5/(x+3) is a one-to-one function, we need to check if every element in the domain corresponds to a unique element in the range. In other words, we need to check if each distinct input (x) produces a distinct output (F(x)).
To do this, we can use the horizontal line test. If any horizontal line intersects the graph of the function at more than one point, then the function is not one-to-one. If every horizontal line intersects the graph at most once, then the function is one-to-one.
Let's first solve the equation for F(x) = y to find the inverse of the function.
F(x) = 5/(x+3)
To find the inverse, we swap x and y:
x = 5/(y+3)
Now, solve for y:
xy + 3x = 5
xy = 5 - 3x
y = (5 - 3x)/x
The inverse function is given by F^(-1)(x) = (5 - 3x)/x.
To determine if F(x) is one-to-one, we can also find the derivative of the function. If the derivative is always positive or always negative, the function is one-to-one. If the derivative changes sign at any point, the function is not one-to-one.
Differentiating F(x) with respect to x:
F'(x) = -5/(x+3)^2
The derivative is negative for all x except x = -3, where it is undefined.
Since F'(x) is always negative (or undefined at x = -3), the function F(x) = 5/(x+3) is a one-to-one function.
Therefore, the function is one-to-one, and the inverse function is F^(-1)(x) = (5 - 3x)/x.