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Question 1169998: The problem I got incorrect:
Find inverse of h(x)=(3/4)x+12
My steps...
1) y = (3/4)x+12
2) (4/3)y = x+12
3) (4/3)y-12 = x
4) (4/3)x-12 = y
5) h^-1(x) = (4/3)x-12
Am told correct answer should be h^-1(x) = (4/3)(x-12).
Could someone explain the property I missed applying?
Thank you.
Found 3 solutions by josgarithmetic, MathTherapy, greenestamps: Answer by josgarithmetic(39617) (Show Source): Answer by MathTherapy(10552) (Show Source): 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)
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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)
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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?
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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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