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avatar+90 

j(x)=2x-3

 

what it \(j({j}^{-1}(x))\)

and what would be \({j}^{-1}(-1)\)

 

Please explain the steps of how you did it and how to do it.

 

Thanks

-Dom

dom6547  Jan 23, 2018
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2+0 Answers

 #1
avatar+765 
+3

Step 1:

  So, to find the inverse, you basically switch x and y, then solve for y.

  j(x) is the same thing as y.

  \(y=2x-3\)

  Switch the x and y.

  \(x=2y-3\)

  Solve for y.

  Add 3 to both sides.

  \(x+3=2y\)

  Divide both sides by 2.

  \(y=\frac{x+3}{2}\)

  Switch the y to j-1(x).

  \(j^{-1}(x)=\frac{x+3}{2}\)

 

Step 2:

  Now for j(j-1(x)).

  You want to plug the value of j-1(x) into j(x) for every x value.

  \(j(\frac{x+3}{2})=2(\frac{x+3}{2})-3\)

  Multiply the 2.

  \(j(j^{-1}(x))=\frac{2(x+3)}{2}-3\)

  The 2's cancel.

  \(j(j^{-1}(x))=x+3-3\)

  The 3's cancel.

  \(j(j^{-1}(x))=x\)

 

Step 3:

  For j-1(-1), plug -1 in for each x value in j-1(x).

  \(j^{-1}(-1)=\frac{(-1)+3}{2}\)

  Add -1 and 3.

  \(j^{-1}(-1)=\frac{2}{2}\)

  2/2 equals 1, so \(j^{-1}(-1)=1\).

AdamTaurus  Jan 23, 2018
 #2
avatar+90 
+1

Thank you!

dom6547  Jan 23, 2018

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