Inverse function question

[Bad Student]

This question comes from MAV 2012 exam 1.

Consider the function



Find, in simplest form, the gradient of the normal to the inverse function at the point where .

[BigAl]

Just interchange x and y then use implicit differentiation

[Bad Student]

Just interchange x and y then use implicit differentiation

Thank you. You are too smart.

[abeybaby]

Since swapping x and y gives the inverse, then dx/dy would give the gradient of the inverse. Therefore, gradient of the normal of the inverse would be -1/(dx/dy) = -dy/dx. So just dofferentiate, sub in e, then put a negative sign in front of your answer

EDIT: look below, I made a mistake

[BasicAcid]

Since swapping x and y gives the inverse, then dx/dy would give the gradient of the inverse. Therefore, gradient of the normal of the inverse would be -1/(dx/dy) = -dy/dx. So just dofferentiate, sub in e, then put a negative sign in front of your answer

When people are able to find shortcuts like these it makes me wonder how much am I currently unaware of...

[abeybaby]

When people are able to find shortcuts like these it makes me wonder how much am I currently unaware of...

maths is incredible, isn't it

[lzxnl]

Since swapping x and y gives the inverse, then dx/dy would give the gradient of the inverse. Therefore, gradient of the normal of the inverse would be -1/(dx/dy) = -dy/dx. So just dofferentiate, sub in e, then put a negative sign in front of your answer

My only qualms with this method would be that although it's very quick and efficient, care needs to be taken to make sure you sub the right value in.
You could go for a slightly slower approach by writing everything out.
y=xe^x => x=ye^y
dx/dy=ye^y+e^y which is the negative of the normal to the inverse
so we want -dx/dy
But we have x=e and we need y. x=e, e=ye^y=>y=1 (you have to just see that)
sub that in, dx/dy=2e
gradient of normal=-2e

If we just went -dy/dx, we would end up with -(xe^x+e^x) which is identical to my answer above replacing x's with y's, but I don't think subbing in x=e is correct here...just my thoughts...

Certainly the solutions agree with my method.

[abeybaby]

My only qualms with this method would be that although it's very quick and efficient, care needs to be taken to make sure you sub the right value in.
You could go for a slightly slower approach by writing everything out.
y=xe^x => x=ye^y
dx/dy=ye^y+e^y which is the negative of the normal to the inverse
so we want -dx/dy
But we have x=e and we need y. x=e, e=ye^y=>y=1 (you have to just see that)
sub that in, dx/dy=2e
gradient of normal=-2e

If we just went -dy/dx, we would end up with -(xe^x+e^x) which is identical to my answer above replacing x's with y's, but I don't think subbing in x=e is correct here...just my thoughts...

Certainly the solutions agree with my method.

Yeah you're right, you wouldn't sub in x=e, you'd have to find the x-value for which y=e, and sub that in. My bad!

[BubbleWrapMan]

Just a few comments:

For this question you have to solve the equation . Equations like this wouldn't normally be solvable, but by inspection we can see that it works when , since the RHS is .

So if you see this sort of equation in an exam 1 (i.e. one you can't solve) then you've either done something wrong or there's an easy value to substitute in and verify.

Of course, you don't always know you have every solution if you simply substitute in values. But in this case is injective (one-to-one), so there is at most one value of for which . Thus is the only solution.

[ahat]

When people are able to find shortcuts like these it makes me wonder how much am I currently unaware of...

This is exactly what I was thinking and makes me fear these sentinels of uncanny ability for taking the same exams as I do.