Help me with this question? (1 Viewer)

[MATHmaster]

Hi everyone this question came to came in our half yearly paper and it kinda hurt me with the wording of the question:

(c) Let y = (4-x)^2 be a function of g(x)

(i) Find the domain where the function g(x) is monotonic increasing
(ii) State the inverse function over this domain
(iii) Sketch the curve in this domain

May anyone help me?

 

[Viscass]

firstly (4-x)^2 can be written as (x-4)^2, which will make it easier to answer the question
i) x>=4 this can be seen visually or you can differntiate the function to find the minumum turning point then find which side is increasing
ii)y=sqrt(x)+4, the domain in g(x) is restricted to x>=4 therefore the range of g^-1(x) is y>=4 which is why we only include the positive 'part' of the root

hope that helped
any confusion please just ask

 

[MATHmaster]

My confusion is that I drew the inverse function y = 4 - sqrt(x) for x>4 in the same domain as the one before... because it said "state the inverse function in this domain". This is why I graphed it for y= 4 - sqrt(x), as I thought the domain was still x>4, passing through (16,0) an open circle at (4, 2)... Is this wrong, because I found the words "this domain" ambiguous

 

[anomalousdecay]

firstly (4-x)^2 can be written as (x-4)^2, which will make it easier to answer the question
i) x>=4 this can be seen visually or you can differntiate the function to find the minumum turning point then find which side is increasing
ii)y=sqrt(x)+4, the domain in g(x) is restricted to x>=4 therefore the range of g^-1(x) is y>=4 which is why we only include the positive 'part' of the root

hope that helped
any confusion please just ask

Sorry, but I don't think it always works if you put (x-4)^2. Find the inverse function of both y=(x-4)^2 and y=(4-x)^2. You should get a different result.