Quick question on inverse functions (1 Viewer)

Jackson94

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Find where the equation y=-x^3 and its inverse intercept, please show working so I can understand how to do it myself, thanks!
 

RealiseNothing

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y = -x^3

It's inverse is when you swap the x and y around so:

x = -y^3

Make y the subject:

y = cube root of -x

Now solving simultaneously:

-x^3 = cube root of -x

-x^9 = -x (Cube both sides)

x^9 = x

x = -1, 0, or 1
 
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SpiralFlex

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y = -x^3

It's inverse is when you swap the x and y around so:

x = -y^3

Make y the subject:

y = cube root of -x

Now solving simultaneously:

-x^3 = cube root of -x

-x^9 = -x (Cube both sides)

x^9 = x

x = -1, 0, or 1

Will wait for Spiral or some one to confirm since I'm not 100% sure this is right.
Lol why are you waiting for me?
 

SpiralFlex

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I interpret the question that is punctuated incorrectly as

Find where the function and its inverse intersect.
 

SpiralFlex

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I would also state in Cartesian coordinate form.
 

Carrotsticks

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The inverse function is drawn by flipping the curve about the line y=x or y=-x.

This means that when it intersects its inverse, it also intersects the line y=x or y=-x.

So all you have to do is solve the pair of simultaneous equations y=-x^3 and y=x or y=-x (depends on the Q) in order to find the inverse intercept.

I have provided a bit of a diagram to help you see it: 1.jpg

Instead of solving simultaneously the equation of the curve and its inverse, you can simply do it with the equation of the curve, and the line y=x or y=-x, since that's where it intersects the inverse anyway.

Here is another example for you to see: 2.jpg

EDIT: Yes, you will have to restrict the domain yourself. You will acquire two solutions, but you cannot have both. Which branch you choose however, will depend on the question.
 

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SpiralFlex

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The inverse function is drawn by flipping the line about the line y=x.

This means that when it intersects its inverse, it also intersects the line y=x.

So all you have to do is solve the pair of simultaneous equations y=-x^3 and y=x in order to find the inverse intercept.

I have provided a bit of a diagram to help you see it: View attachment 24028

Instead of solving simultaneously the equation of the curve and its inverse, you can simply do it with the equation of the curve, and the line y=x, since that's where it intersects the inverse anyway.

Here is another example for you to see: View attachment 24029
The second example is not an inverse function unless you restrict it.
 

Carrotsticks

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The second example is not an inverse function unless you restrict it.
Yes I know it is not a function by definition due to the unrestricted domain, but it's so he can properly see the action of the 'flipping', and how it yields two solutions. If you restrict the domain beforehand, then you may miss out on solutions.

In exams, I draw the whole thing unrestricted (on the edge of the page just for myself to see), THEN I restrict the domain according to the question (if it specifies which branch. If it doesn't, then I write down the 2 possible cases)
 
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largarithmic

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I think "find the intersection of a curve and its inverse" and "find the intersection of a function and its inverse function" are two pretty different questions... in particular the former is much nastier. try find the intersection of y = 10x^3 - 10x for instance and its inverse x = 10y^3 - y. You'll give yourself a serious headache (there are nine solutions!)
 

Carrotsticks

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I think "find the intersection of a curve and its inverse" and "find the intersection of a function and its inverse function" are two pretty different questions... in particular the former is much nastier. try find the intersection of y = 10x^3 - 10x for instance and its inverse x = 10y^3 - y. You'll give yourself a serious headache (there are nine solutions!)
lol but this is the HSC. Asif they will expect that of a student.

If you put the two curves you mentioned (with unrestricted domain) on a graphing program, you will see a 2x2 grid sorta thing. It would be a very interesting question to try to find the area of that '2x2' grid.
 

largarithmic

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lol but this is the HSC. Asif they will expect that of a student.

If you put the two curves you mentioned (with unrestricted domain) on a graphing program, you will see a 2x2 grid sorta thing. It would be a very interesting question to try to find the area of that '2x2' grid.
oooh thats legit... Id imagine you could cut it up and integrate it?

alsoooooo this reminds me, you know theres an awesome theorem about the intersection of two cubics? http://en.wikipedia.org/wiki/Cayley–Bacharach_theorem

lots of random geometric results are made almost trivial by that awesome theorem - you can treat things like a triple of lines, for instance, as a single cubic eqn e.g. (x+y-1)(x+2y-3)(x-4y+5) = 0 is a set of 3 lines, or a circle and a line e.g. (x^2+y^2-1)(x+y-1)=0. For example http://en.wikipedia.org/wiki/Pivot_theorem is a special case
 

Carrotsticks

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oooh thats legit... Id imagine you could cut it up and integrate it?

alsoooooo this reminds me, you know theres an awesome theorem about the intersection of two cubics? http://en.wikipedia.org/wiki/Cayley–Bacharach_theorem

lots of random geometric results are made almost trivial by that awesome theorem - you can treat things like a triple of lines, for instance, as a single cubic eqn e.g. (x+y-1)(x+2y-3)(x-4y+5) = 0 is a set of 3 lines, or a circle and a line e.g. (x^2+y^2-1)(x+y-1)=0. For example http://en.wikipedia.org/wiki/Pivot_theorem is a special case
I've run across that theorem before when I randomly browse Maths articles on Wikipedia (I know it sounds sad, but it is actually quite interesting!). It would be interesting to have an elementary proof of it appear in the HSC.

Also regarding the area problem, there must be an easier way. The cutting up and integrating is too... brute force. Not elegant.
 

largarithmic

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I've run across that theorem before when I randomly browse Maths articles on Wikipedia (I know it sounds sad, but it is actually quite interesting!). It would be interesting to have an elementary proof of it appear in the HSC.
reading maths articles on wikipedia is never a bad thing! I heard the guy who wrote the q8 this year based it on a theorem he found on wikipedia :p and I doubt theres an elementary proof of that theorem, Ive heard its really hard, and only works for cubics (has no quadratic/quartic analogue) which is weird
 

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