+101 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
+895 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 \(j^{-1}(-1)=1\).
