FNGR 2 | Lesson 5 | Making Connections (Inverse Operations)

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Inverse Operations

When you perform the indicated operation on inverses, the result is the identity of that operation.  For this reason, we say that you perform the inverse to “undo” the operation.   

Example 1:  Addition

The identity for addition is 0 since when you add 0 to any number, you get that number as the result, as shown below:

\(a+0 = a\)  or  \(0+a=a\)

The inverse of addition is subtraction or to add the opposite.  When you add inverses (opposites), the result is the identity, as shown below:

\(a+(-a)=0\)


Example 2:  Multiplication

The identity for addition is 1 since when you multiply any number by 1, you get that number as the result, as shown below:

\(a\times 1 = a\)  or  \(1\times a = a\)

The inverse of multiplication is division or to multiply by the reciprocal.  When you multiply inverses (reciprocal), the result is the identity, as shown below:

\(a\times \dfrac1a = 1\)  or  \(\dfrac 1a \times a = 1\)


Example 3:  Composition of Functions

The identity for functions is  \(y=x\)  or  \(f(x) = x\)  since when you input any number, you get that number as the result, as shown below:

\(f(a)=a\)

The inverse operation of a function is composition. When you compose inverse functions, the result is the identity, as shown below:

\(f(f^{-1}(x))=x\)  and  \(f^{-1}(f(x))=x\)

Example:  Given \(f(x) = -2x-5\) and \(g(x)=-\dfrac x2-\dfrac 52\) , find \(f(g(x))\) and \(g(f(x))\) . 

\( \begin{equation} \begin{aligned} f(g(x)) & = f(-\dfrac x2-\dfrac 52)  \\\\ & =-2(-\dfrac x2-\dfrac 52)-5 \\\\ & =x+5-5 \\\\ & =x\end{aligned} \end{equation}\)

\( \begin{equation} \begin{aligned} g(f(x)) & = g(-2x-5) \\ \\ &=-\dfrac {(-2x-5)}2-\dfrac 52 \\ \\ &=\dfrac {2x+5}2-\dfrac 52 \\\\&=x+\dfrac52 -\dfrac 52 \\\\& =x\end{aligned} \end{equation}\)

Additionally, if \(f(a)=b\) , then \(f^{-1}(b)=a\) .

Example: If \(f(x)=2x-1\)  and  \(f(3)=5\) , then for some function \(f\) , \(f^{-1}(5) = 3\) .


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Go to Try This! (Writing Inverse Equations)