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A Value of the Inverse

Part of the Functions & Transformations section of Coddy's Algebra 2 journey. Lesson 28 of 55.

You do not need the whole inverse rule to find one value of it. g−1(23){g}^{-1}\left(23\right) asks: which input does g send to 23? So write g(x)=23g\left(x\right) = 23 and solve for x.

Take g(x)=3x+2g\left(x\right) = 3x+2. Using the rule turns the question into 3x+2=233x+2 = 23, so 3x=213x = 21 and x is 7. That means g−1(23)=7{g}^{-1}\left(23\right) = 7. Check forwards: g(7) is 21 + 2, which is 23.

This works because the inverse swaps inputs and outputs: g sends 7 to 23 exactly when its inverse sends 23 to 7. Solving one equation is quicker than finding the full inverse rule and then evaluating it.

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Challenge

For f(x)=4x−5f\left(x\right) = 4x-5, find f−1(15){f}^{-1}\left(15\right). The board holds f(x)=15f\left(x\right) = 15, which asks for the input that f sends to 15. Solve it.

Try it yourself

f(x)=15f\left(x\right) = 15

Get xx alone

Givenf(x)=4x−5f\left(x\right) = 4x-5

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