Can someone rephrase this with different equations please I really need this!
Even if the inverses of both g(x) and f(x) are functions, the inverse of the sum of g(x) and f(x) may not be a function. As an example, I will use f(x) = 3x +2 and g(x) = 2x - 3. Both are one-to-one functions. The sum of f(x) and g(x) is 5x - 1. As 5x - 1 is also a one-to-one function, its inverse is a function as well. This example is in favor of the sum of two one-to-one function also being a one-to-one function. But take c(x) = (5x - 1)/ ((x/2) + 4 ) as another example. C(x) + f(x) is still a function, but the inverse of c(x) + f(x) is not. The inverse form does not pass the vertical line test in desmos. So, not every sum of two one-to-one functions is also a one-to-one function.
idk
Sure! Here's a rephrased version with different equations:
Even if the inverses of functions h(x) and j(x) exist, the inverse of the sum of h(x) and j(x) may not be a function. Let's consider h(x) = 2x + 1 and j(x) = 3x - 4. Both of these functions are one-to-one, meaning each input has a unique output.
The sum of h(x) and j(x) is k(x) = 5x - 3. Since k(x) is also a one-to-one function, its inverse exists and is also a function. So, in this example, the sum of two one-to-one functions results in another one-to-one function.
Now, let's consider a new function m(x) = (5x - 3)/((x/2) + 5). If we add m(x) to h(x), we get the function n(x) = (5x - 3)/((x/2) + 5) + 2x + 1. While n(x) is still a function, its inverse is not a function. This can be observed by using a graphing tool like Desmos and applying the vertical line test. The inverse of n(x) does not pass the test, indicating that it is not a one-to-one function.
Therefore, it is not always the case that the sum of two one-to-one functions will also be a one-to-one function.