![\"Domain](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200955\/CNX_Precalc_Figure_01_07_0032.jpg\")
Figure 3.<\/b> Domain and range of a function and its inverse[\/caption]\n<\/p>The outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]Figure 3.<\/b> Domain and range of a function and its inverse[\/caption]\n<\/p>
When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] is [latex]{f}^{-1}\\left(x\\right)={x}^{2}[\/latex], because a square \"undoes\" a square root; but the square is only the inverse of the square root on the domain [latex]\\left[0,\\infty \\right)[\/latex], since that is the range of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex].<\/p>\n
We can look at this problem from the other side, starting with the square (toolkit quadratic) function [latex]f\\left(x\\right)={x}^{2}[\/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and \u20133. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the \"inverse\" is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.<\/p>\n
In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).<\/p>\n
If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}[\/latex] on [latex]\\left[1,\\infty \\right)[\/latex], then the inverse function is [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+1[\/latex].<\/p>\n\n
Q & A<\/strong><\/p>\nIs it possible for a function to have more than one inverse?<\/strong>\n No. If two supposedly different functions, say, [latex]g[\/latex] and [latex]h[\/latex], both meet the definition of being inverses of another function [latex]f[\/latex], then you can prove that [latex]g=h[\/latex]. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.<\/em><\/p>\n\n<\/div>\n The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n\n<\/div>\n Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed below. We restrict the domain in such a fashion that the function assumes all y<\/em>-values exactly once.<\/p>\n\n The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.<\/p>\n The absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.<\/p>\n The reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n\n<\/div>\n\nA General Note: Domain and Range of Inverse Functions<\/h3>\n
\nHow To: Given a function, find the domain and range of its inverse.\n<\/strong><\/h3>\n
\n\n\nExample 4: Finding the Inverses of Toolkit Functions<\/h3>\n
Constant<\/td>\n Identity<\/td>\n Quadratic<\/td>\n Cubic<\/td>\n Reciprocal<\/td>\n<\/tr> [latex]f\\left(x\\right)=c[\/latex]<\/td>\n [latex]f\\left(x\\right)=x[\/latex]<\/td>\n [latex]f\\left(x\\right)={x}^{2}[\/latex]<\/td>\n [latex]f\\left(x\\right)={x}^{3}[\/latex]<\/td>\n [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/td>\n<\/tr> Reciprocal squared<\/td>\n Cube root<\/td>\n Square root<\/td>\n Absolute value<\/td>\n [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]<\/td>\n [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]<\/td>\n [latex]f\\left(x\\right)=\\sqrt{x}[\/latex]<\/td>\n [latex]f\\left(x\\right)=|x|[\/latex]<\/td>\n \nSolution<\/h3>\n
Analysis of the Solution<\/h3>\nWe can see that these functions (if unrestricted) are not one-to-one by looking at their graphs.\u00a0They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200956\/CNX_Precalc_Figure_01_07_004ab2.jpg\")
Figure 4.<\/b> (a) Absolute value (b) Reciprocal squared[\/caption]\n\n<\/div>\n<\/div>\n<\/div>\n\n
Try It 4<\/h3>\n
The domain of function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.<\/p>\nSolution<\/a>\n\n<\/div>","rendered":"![\"Domain](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200955\/CNX_Precalc_Figure_01_07_0032.jpg\")
<\/p>\n
The outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.<\/p>\n
Figure 3.<\/b> Domain and range of a function and its inverse<\/p>\n<\/div>\n
When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] is [latex]{f}^{-1}\\left(x\\right)={x}^{2}[\/latex], because a square “undoes” a square root; but the square is only the inverse of the square root on the domain [latex]\\left[0,\\infty \\right)[\/latex], since that is the range of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex].<\/p>\n
We can look at this problem from the other side, starting with the square (toolkit quadratic) function [latex]f\\left(x\\right)={x}^{2}[\/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and \u20133. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.<\/p>\n
In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).<\/p>\n
If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}[\/latex] on [latex]\\left[1,\\infty \\right)[\/latex], then the inverse function is [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+1[\/latex].<\/p>\n
\n- The domain of [latex]f[\/latex] = range of [latex]{f}^{-1}[\/latex] = [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\n
- The domain of [latex]{f}^{-1}[\/latex] = range of [latex]f[\/latex] = [latex]\\left[0,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n
\n
Q & A<\/strong><\/p>\n
Is it possible for a function to have more than one inverse?<\/strong><\/p>\n
No. If two supposedly different functions, say, [latex]g[\/latex] and [latex]h[\/latex], both meet the definition of being inverses of another function [latex]f[\/latex], then you can prove that [latex]g=h[\/latex]. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.<\/em><\/p>\n<\/div>\n\n\n
A General Note: Domain and Range of Inverse Functions<\/h3>\n
The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n
The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<\/div>\n
How To: Given a function, find the domain and range of its inverse.
\n<\/strong><\/h3>\n
\n- If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.<\/li>\n
- If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.<\/li>\n<\/ol>\n<\/div>\n
\n\n\n
Example 4: Finding the Inverses of Toolkit Functions<\/h3>\n
Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed below. We restrict the domain in such a fashion that the function assumes all y<\/em>-values exactly once.<\/p>\n\n\n \n\nConstant<\/td>\n Identity<\/td>\n Quadratic<\/td>\n Cubic<\/td>\n Reciprocal<\/td>\n<\/tr>\n \n[latex]f\\left(x\\right)=c[\/latex]<\/td>\n [latex]f\\left(x\\right)=x[\/latex]<\/td>\n [latex]f\\left(x\\right)={x}^{2}[\/latex]<\/td>\n [latex]f\\left(x\\right)={x}^{3}[\/latex]<\/td>\n [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/td>\n<\/tr>\n \nReciprocal squared<\/td>\n Cube root<\/td>\n Square root<\/td>\n Absolute value<\/td>\n \n<\/td>\n<\/tr>\n \n[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]<\/td>\n [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]<\/td>\n [latex]f\\left(x\\right)=\\sqrt{x}[\/latex]<\/td>\n [latex]f\\left(x\\right)=|x|[\/latex]<\/td>\n \n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n\n![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200956\/CNX_Precalc_Figure_01_07_004ab2.jpg\")
<\/p>\n
Solution<\/h3>\n
The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.<\/p>\n
The absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.<\/p>\n
The reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<\/div>\n
Analysis of the Solution<\/h3>\n
We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs.\u00a0They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.<\/p>\n
Figure 4.<\/b> (a) Absolute value (b) Reciprocal squared<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n
Try It 4<\/h3>\n
The domain of function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.<\/p>\n
Solution<\/a><\/p>\n<\/div>\n\n\t\t\t\n\t\t\t
Candela Citations<\/h3>\n\t\t\t\t\t\n\t\t\t\t\t\t \n\t\t\t\t\t\t\t CC licensed content, Shared previously<\/div>- Precalculus. Authored by<\/strong>: Jay Abramson, et al.. Provided by<\/strong>: OpenStax. Located at<\/strong>: http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. License<\/strong>: CC BY: Attribution<\/a><\/em>. License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1029","chapter","type-chapter","status-publish","hentry"],"part":1022,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1029","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1029\/revisions"}],"predecessor-version":[{"id":2466,"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1029\/revisions\/2466"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1022"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1029\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/wp\/v2\/media?parent=1029"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1029"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1029"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/wp-json\/wp\/v2\/license?post=1029"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}