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Inverses and Radical Functions

Learn how to find the inverse of a polynomial function by restricting its domain. This lesson in College Algebra covers inverse functions and radical functions with practical examples to help you understand and apply the concepts.

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    College Algebra Inverses and Radical Functions Restrict the domain to find the inverse of a polynomial function Sc far, we have been able t o find the inverse functions c f cubic functions without having to restrict their domains. However, as we know, not all cubit polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one. but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to- one. we must restrict their domain in order to find their inverses. A General Note: Restricting the Domain I f a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one. thus creating a new function, this new function will have an inverse. How Tc: Given a polynomial function, restrict the domain of = function that is not one-to-one and then find the inverse. Restrict the domain by determining a domain on which the original function is one-to-one. Replace flyl with y. niterchange x and y. Solve fory and rename the function or pair of function / Revise the formula for / by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function. Example 3: Restricting the Domain to Find the nverse o f a Polynomial Function Find the inverse function of it / (*) = ( * 4) 1 , x > 4 / ( r ) = (u: 4) i , a : < 4 Solution The original function / ( « ) = ( » 4) 4 is not one-to-one, but the function is restricted to a domain o f i > 4 Figure 5

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    To find the inverse, start by replacing f (®) with the simple variable y. y = ( x 4) Interchange -r and y. x — ( y 4) Take the square root. .Lii/x — y 4 A d d 4 t o both sides. I 4 1 y<r - y This is not a function as written. We need to examine the restrictions o n the domain o f the original function to determine the inverse. Since we reversed the roles of x and y for the original fix), we looked at the domain: the values x could assume. When we reversed the roles o f x ar d y, this gave us the values y could assume. For this function. z > 4 , so for the inverse, we should have Jf > 4 , which is what our inverse function gives. The domain of the original function was restricted t o z > 4 , so the outputs of the inverse need to be the same, / > 4 . and we must use the - case: / * (x) = 4 I The domain of the original function was restricted to z < 4 , so the outputs of the inverse need to be the same, / W < 4 , and we must use the - case: / 1 (x) = 4 Analysis of the Solution On the graphs below, we see the original function graphed o n the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line 7 = 2 - . The coordinate pair (4,0) is on the graph of f and the coordinate pair ( 0 , 4 ) is o n the graph of / 1 . For any coordinate pair, if [a, t j is o n the graph of f then (ir. tr) is on the graph of / 1 . Finally, observe that the graph o f f intersects the graph of / 1 o r the line y = x. Points of intersection for the graphs of f and / 1

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    Example 4: Finding the Inverse of a Quadratic Function When the Restriction Is N o t Specified Restrict the domain and then find the inverse of / ( » ) = ( * 2) 1 3 Solution We can see this is a parabola with vertex at (2, 3) that opens upward. Because the graph will be decreasing on one side o f the vertex and increasing o n the other side, we can restrict this function to a domain o n which it will be one-to-one by limiting the domain t o x > 2 To find the inverse, we will use the vertex form of the quadratic. We start by replacing fbj with a simple variable, y, then solve forx y = ( x 2 ) 2 3 lutcrcluiiigi: x and y. X = 2)* 3 A d d 3 to both sides. IX I 3 = ( y 2) 2 Take the square root. 1 J - / x I 3 = j 2 Add 2 to b o t h sides. 2 1 v'x I 3 y Rename the lurictioti. J 1 (x) = 2 1 Now we need t o determine which case t o use. Because we restricted our original function to a domain o f x > 2 , the outputs of the inverse should be the same, telling us t o utilize the + case / 1 ( x ) = 2 I I f the quadratic had not been given i n vertex form, rewriting it into vertex form would he the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain. Analysis o f the Solution Notice that we arbitrarily decided to restrict the domain o n x > 2 . We could just have easily opted to restrict the domain o n x < 2 , i n which case

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