FNGR 2 | Lesson 5 | Try This (Inverse Relationships)


Inverse Relationships

In Exploring Function Reflections, when two functions are reflected over the line \(y=x\) , then the point \((a,b)\) from one function and the point \((b,a)\) from the second function form a line segment perpendicular to the line \(y=x\) and the points are equidistant to the line \(y=x\).   

If both curves are functions (they pass the vertical line test), then this relationship is defined as the inverse of two functions and they are reflected over the line \(y=x\) which is the identity function.  (Note: identity means that for any input \(x\), the output is \(x\) or \(f(x)=x\))

For example, if the point \((1,2)\) lies on the function \(f(x)\) , then \((2,1)\) lies on the inverse of \(f(x)\) .  We write the inverse function as  \(f^{-1}(x)\) .  Note that the inverse of a function is not its reciprocal; therefore, \(f^{-1}(x) \neq \dfrac{1}{f(x)}\).

Also, if a function has an inverse, then it is said to be 1-to-1.  A function \(f\) is 1-to-1 if no two elements in the domain of \(f\) correspond to the same element in the range of \(f\) . In other words, each \(x\) in the domain has exactly one image in the range.  

Finally, a function that passes the horizontal line test (when horizontal lines do not intersect the graph of the function \(f\) in more than one point), then the function is 1-to-1.

Determine which of the following are inverse functions and explain your reasoning.

1.  \(f(x) = -2x~–~5\) and \(g(x) = -\dfrac x2 ~-~ \dfrac 52\)

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2.  \(h(x) = x + 5\)  and \(k(x)=-x-5\)

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3.  \(r(x)=x^3\) and \(t(x)=\sqrt[3]{x}\)

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Check your solutions here.


Go to Watch (Graphing Inverse Functions)