Never "trust" any answer the other person gives. In MOST cases, his answers are WRONG!! So, I sympathize with you getting a wrong

answer after determining the inverse, but also getting another wrong answer from someone who claims to be a tutor!

Just "throw" his answer in the GARBAGE, or SIMPLY do this: The inverse of h  is  



Inverse of 

Step 1):  <==== Correct

Step 2):  ----- Switching variables

Step 3): 

Step 4):  

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!
 


Between your steps (1) and (2), you attempted to multiply everything by 4/3, but you did not -- you didn't multiply the "12" by 4/3.  Your step (2) should show


(4/3)y=x+16


That would lead you to the answer


h^-1(x) = (4/3)x-16


which is equivalent to the given answer of


h^-1(x) = (4/3)(x-12)


------------------------------------


Note for many simple functions, there is an easier and faster way to find the inverse of a function, using the idea that the inverse "un-does" what the function does.


In this example, what the function does to the input is

(1) multiply by (3/4); and

(2) add 12


To "undo" those operations, the inverse function has to perform the opposite operations in the opposite order:

(1) subtract 12; and

(2) divide by (3/4) -- i.e., multiply by (4/3)


Those two steps immediately give you the inverse function:


h^-1(x) = (4/3)(x-12)


-----------------------------------------------


comment from student....


Thank you for suggestion on how to determine inverse of many simple functions; that appears a sound approach. But going back to my initial mistake, let’s consider only the part of this problem where we start with y isolated on one side ( y= (3/4)x+12 ), and we want to rearrange the equation so that instead x is the isolated variable. When I undo the multiplication of (3/4)x on the right side by dividing y on the left by 3/4 (actually multiplying by inverse), at that point, y is the only term on the left side. When proceeding by moving the 12 over to left I’m puzzled how I should have known to multiply it by 4/3 also (or that I could have stopped short of doing the final distribution by writing 4/3(x-12)). Is there something incorrect/unsound with beginning to rearrange the equation by undoing the multiplication? 


----------------------------------------------------


No; there is nothing wrong with "undoing the multiplication" to start on the problem.  But you need to do it correctly.


Undoing the multiplication means multiplying EVERYTHING on both sides of the equation by (4/3).  In your work, you only multiplied parts of the equation by (4/3).




--> 

--> 


When you "proceed" from there, there is no "12" to move to the left; it is a "16". 

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