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Inverse graph question (1 Viewer)

scaryshark09

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the answer is D but i got get how
like take y=x^2 as an example. the function and the inverse would intersect at x=0 and x=1
at x=0 the tangents would be perpendicular, right??
 

scaryshark09

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can someone also please explain this?
 

carrotsss

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can someone also please explain this?
- The derivative can never be negative because siny>=-1. This makes C impossible
- The derivative can never be greater than 2. This makes A impossible
- The derivative approaches 0, at which point the y value cannot increase further, so it can’t just go up forever. This makes D impossible
Hence, B
 
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the answer is D but i got get how
like take y=x^2 as an example. the function and the inverse would intersect at x=0 and x=1
at x=0 the tangents would be perpendicular, right??
for this question i basically just made sample graphs lol

for y=x^2, if take the inverse u get y=sqrt x (assuming u take the positive), diffing it gives u 1/(2sqrt x), subbing x=1 gives a gradient of 1/2.
now if u diff y=x^2, u get y=2x, subbing in x=1 u get y=2.

Gradient of 2 and 1/2 is not perp, it needs to be the negative reciprocal.
 
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can someone also please explain this?
this question worst case scenario u can just use ur calc and test points. U realise that as y-> a big value, dy/dx -> 0+. U can immediately rule out a,c, d. But if u want the 'proper' way, carrots gives a great definition. (this is just to point out that in an exam u can still manage the get the mark, despite not being 100% sure :) )
 

Luukas.2

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the answer is D but i got get how
like take y=x^2 as an example. the function and the inverse would intersect at x=0 and x=1
at x=0 the tangents would be perpendicular, right??
A function and its inverse must meet on y = x, but if a function is its own inverse (like for xy = 1 or y = -x) it will meet itself at every point in its domain.

f(x) = -x is a function that meets all the criteria specified and it is clear for it that (A), (B), and (C) are false.
 

yanujw

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can someone also please explain this?
Someone explained a good way of doing it by deduction above. Another way (which I wish I realised when I actually did this question in the HSC) is to seperate the differential equation and find . It's not as hard as it looks - there is a way to do this using an extension 2 technique, as well as a way with extension 1 theory alone, which I'll leave as an exercise to you.
 

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