Inverse graph question (1 Viewer)

[scaryshark09]

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]

can someone also please explain this?

 

[carrotsss]

View attachment 40688
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

 

[iheartcookies1012]

View attachment 40685
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.

 

[iheartcookies1012]

View attachment 40688
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]

View attachment 40685
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]

View attachment 40688
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.