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# A question for you

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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
#1
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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.

$$x+3=2y$$

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).

$$j^{-1}(-1)=\frac{(-1)+3}{2}$$

$$j^{-1}(-1)=\frac{2}{2}$$

2/2 equals 1, so $$j^{-1}(-1)=1$$.

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Thank you!

dom6547  Jan 23, 2018