y = x2 where the domain is the set of real numbers does not have an inverse, because the square root function is a one-two-two mapping (except at 0).
Any polynomial with more than one root, over the reals, has no inverse.
y = 1/x has no inverse across 0.
But it is possible to define the domain so that each of these functions has an inverse.
For example y = x2 where x is non-negative has the square root function as its inverse.
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If f(x)=y, then the inverse function solves for y when x=f(y). You may have to restrict the domain for the inverse function to be a function. Use this concept when finding the inverse of hyperbolic functions.
The only trig functions i can think of with horizontal assymptotes are the inverse trig functions. and they go assymptotic for everytime the non-inverse function is equal to zero.
When graphing functions, an inverse function will be symmetric to the original function about the line y = x. Since a constant function is simply a straight, horizontal line, its inverse would be a straight, vertical line. However, a vertical line is not a function. Therefore, constant functions do not have inverse functions. Another way of figuring this question can be achieved using the horizontal line test. Look at your original function on a graph. If any horizontal line intersects the graph of the original function more than once, the original function does not have an inverse. The constant function is a horizontal line. Under the assumptions of the horizontal line test, a horizontal line infinitely will cross the original function. Thus, the constant function does not have an inverse function.
the output is divided by 4
The inverse of the cosine is the secant.