Inverse Functions -_- (1 Viewer)

[scardizzle]

This Q comes from the 1996 HSC paper

Consider the function f(x) = 1/4 [(x-1)^2 + 7]

Rigmarole:

(i)sketch the parabola

(ii) what is the largest domain containing x = 3 that has an inverse function?

(iii)sketch y = f^-1(x)

Now here's what i dont understand:

(iv) Let a be a real number not in the domain found in part (ii). Find f^-1(f(a))

answers say 2-a

any help would be much appreciated

 

[scardizzle]

no takers?

 

[yibbon]

This Q comes from the 1996 HSC paper

Consider the function f(x) = 1/4 [(x-1)^2 + 7]

Rigmarole:

(i)sketch the parabola

(ii) what is the largest domain containing x = 3 that has an inverse function?

(iii)sketch y = f^-1(x)

Now here's what i dont understand:

(iv) Let a be a real number not in the domain found in part (ii). Find f^-1(f(a))

answers say 2-a

any help would be much appreciated

To solve you find f^-1(x) which is y = 1-root(4x-7), then sub f(a) into the x value there (1/4 and 4 cross out in the process):

f(a) = [(a-1)^2]/4 + 7

f^-1(f(a)) = 1-root(-7+7+(a-1)^2)
f^-1(f(a)) = 1-(a-1)
f^-1(f(a)) = 2-a

 

[Aquawhite]

Above post seems a good solve...

This question would be worth a maximum of 2 marks in the exam... I'd say 1. Don't panic too much. lol

 

[oly1991]

To solve you find f^-1(x) which is y = 1-root(4x-7), then sub f(a) into the x value there (1/4 and 4 cross out in the process):

f(a) = [(a-1)^2]/4 + 7

f^-1(f(a)) = 1-root(-7+7+(a-1)^2)
f^-1(f(a)) = 1-(a-1)
f^-1(f(a)) = 2-a

lol i kept doing this and i kept getting just (-a) and i could not see were i went wrong.....i know realise that i didnt put the (a-1) in brackets in the 2nd last step haha silly me.

 

[scardizzle]

To solve you find f^-1(x) which is y = 1-root(4x-7), then sub f(a) into the x value there (1/4 and 4 cross out in the process):

f(a) = [(a-1)^2]/4 + 7

f^-1(f(a)) = 1-root(-7+7+(a-1)^2)
f^-1(f(a)) = 1-(a-1)
f^-1(f(a)) = 2-a

isn't the inverse function y = 1 + root(4x - 7) ? why do you take the negative root?

Above post seems a good solve...

This question would be worth a maximum of 2 marks in the exam... I'd say 1. Don't panic too much. lol

I was afraid i was missing some important formula or something the solutions say as the first step of working :
If a<1 then f(a) = f(1+(1-a)) = f(2-a)
since x = 1 is the axis of symmetry

anyone care to enlighten me?

EDIT: oh i think i get it now since its a parabola, a which lies outside the domain has a y value inside the range since its an even function anyone care to confirm?

 

[yibbon]

isn't the inverse function y = 1 + root(4x - 7) ? why do you take the negative root?

EDIT: Hmm.

Ah I remember: a is taken outside the normal domain (ie not on the curve in our graph). Hence y <= 1, otherwise it would be on the curve at y >=1.

Theres a graph of the inverse

http://www20.wolframalpha.com/input/?i=y=1+sqrt(4x-7)

 

[scardizzle]

ah okay i understand now - very smart