{"id":1037,"date":"2015-11-12T18:35:32","date_gmt":"2015-11-12T18:35:32","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1037"},"modified":"2015-11-12T18:35:32","modified_gmt":"2015-11-12T18:35:32","slug":"use-the-graph-of-a-function-to-graph-its-inverse","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/use-the-graph-of-a-function-to-graph-its-inverse\/","title":{"raw":"Use the graph of a function to graph its inverse","rendered":"Use the graph of a function to graph its inverse"},"content":{"raw":"

Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as in Figure 7.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"GraphFigure 7.<\/b> Quadratic function with domain restricted to [0, \u221e).[\/caption]\n<\/p>

Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.<\/p>\n

We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]<\/p>\nWe notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown in Figure 8.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"GraphFigure 8.<\/b> Square and square-root functions on the non-negative domain[\/caption]\n

This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.<\/p>\n\n

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Example 10: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\nGiven the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\right)[\/latex].\n<\/span>\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"GraphFigure 9<\/b>[\/caption]\n\n<\/div>\n
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Solution<\/h3>\n

This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\nIf we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in Figure 10.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"GraphFigure 10.<\/b> The function and its inverse, showing reflection about the identity line[\/caption]\n\n<\/div>\n<\/div>\n<\/div>\n

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Try It 9<\/h3>\n

Draw graphs of the functions [latex]f\\text{ }[\/latex] and [latex]\\text{ }{f}^{-1}[\/latex].<\/p>\nSolution<\/a>\n\n<\/div>\n

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Q & A <\/strong><\/p>\nIs there any function that is equal to its own inverse?<\/strong>\n

Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em><\/p>\n\n

[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/div>\n

Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\n\n<\/div>\n<\/div>","rendered":"

Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as in Figure 7.<\/p>\n

\"Graph<\/p>\n

Figure 7.<\/b> Quadratic function with domain restricted to [0, \u221e).<\/p>\n<\/div>\n

Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.<\/p>\n

We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]<\/p>\n

We notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown in Figure 8.<\/p>\n

\"Graph<\/p>\n

Figure 8.<\/b> Square and square-root functions on the non-negative domain<\/p>\n<\/div>\n

This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.<\/p>\n

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Example 10: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\n

Given the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\right)[\/latex].
\n<\/span><\/p>\n

\"Graph<\/p>\n

Figure 9<\/b><\/p>\n<\/div>\n<\/div>\n

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Solution<\/h3>\n

This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n

If we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in Figure 10.<\/p>\n

\"Graph<\/p>\n

Figure 10.<\/b> The function and its inverse, showing reflection about the identity line<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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Try It 9<\/h3>\n

Draw graphs of the functions [latex]f\\text{ }[\/latex] and [latex]\\text{ }{f}^{-1}[\/latex].<\/p>\n

Solution<\/a><\/p>\n<\/div>\n

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Q & A <\/strong><\/p>\n

Is there any function that is equal to its own inverse?<\/strong><\/p>\n

Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em><\/p>\n

[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/div>\n

Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\n<\/div>\n<\/div>\n\n\t\t\t

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