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Question 1

a) Find the inverse of y = sqrt(x-2) + 1,

b) State the domain and the range of y-1

Question 2

  • f(x) = sqrt(x-1)
  • g(x) = x + (1/x)

Find f(g(x)) and g(f(x))

Question 3 Graph for Maths Methods question

Find a, b, c and d in that graph equation.

Question 4

Find the 'tan' value of the acute angle between the lines 3x+4y=6 and x-2y=2. Also find the equation of the line which passes perpendicular to the first line and through the mid-point of the line connecting the point of intersection of those two lines and (-4,6)

Question 5

Factorise in real domain: 2x2 - x - x (b2-4ac gives negative result!)

The questions aren't just those or in that order, but they are the ones I have trouble with.

Can you help me please?

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2 Answers

3

Question 1

a) Find the inverse of y = sqrt(x-2) + 1,

Inverse: (swap y and x)

  1. x = sqrt(y-1) + 1
  2. x-1 = sqrt(y-1)
  3. (x-1)2) = y-1
  4. y = (x-1)2 +1

The rest was algebra :)

b) State the domain and the range of y-1

To approach this, you should look at the original function: y = sqrt(x-2) + 1.

  • dom y: x ≥ 2
  • ran y: y ≥ 1

For the inverse, the domains and ranges are just swapped around.

  • dom y-1: x ≥ 1
  • ran y-1: y ≥ 2

Note that if you just tried to work out the domain and range from the formula you found in part (a), you can get the incorrect answer, because it's not taking into account that sqrt(x-2) + 1 only shows one half of the parabola.

Please comment if you don't know what I mean.

Question 2

  • f(x) = sqrt(x-1)
  • g(x) = x + (1/x)

To find f(g(x)), simply replace all x in f(x) with g(x):

  1. f(g(x)) = sqrt(g(x)-1)
  2. f(g(x)) = sqrt(x + (1/x) -1)

Similar idea for g(f(x)):

  1. g(f(x)) = f(x) + 1/f(x)
  2. g(f(x)) = sqrt(x-1) + 1/sqrt(x-1)

Question 3

I assume you meant: y = a sqrt(x+b) + c

  • We have 3 unknowns to work out (a, b and c)

And three pieces of information (hence we can solve):

  1. The graph 'begins' at x = -1 (all square root graphs 'begin' somewhere)
  2. The graph goes through (-1, -2)
  3. The graph goes through (0, 0)

Our first piece of information tells us that b = 1. (x+b) must be greater than or equal to zero because it's inside the square root. Since the graph begins at x = -1, then b = 1 allows (x+b) = 0, the minimum possible value for the square root (and hence where the graph 'begins')

Our second and third pieces of information can be substituted into our equation: y = a sqrt(x+1) + c (we've substituted in b = 1)

  1. -2 = a sqrt(-1 + 1) + c, which gives: c = -2 (since the square root term equals zero)
  2. 0 = a sqrt(0 + 1) + c, which gives: a = 2 (substitute c = -2, the rest is algebra)

So with a = 2, b = 1, and c = -2, we get:

  • y = 2sqrt(x+1) - 2

Edit: Added in Question 2 and Question 3 now. Hope these helped explain the thought process. Please comment if you've still got any more problems -- otherwise, you can accept my answer by selecting the tick, if you like :)

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1 
 thank you very much. I understand all until "because it's not taking into account that sqrt(x-2) + 1 only shows one half of the parabola." is that because sqrt's can be + and -? btw, how do i give out bounty, does it have to be before any answers come in? – Impractical Dec 23 at 23:12
Spot on about the domain. Bounty can only be offered if the question has no accepted answer. – Collin Li♦ Dec 24 at 2:42
Also, the question needs to be 24 hours old or something like that: merspi.com/faq#bounty – Collin Li♦ Dec 24 at 7:17
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Question 1

a) y = (x+1)2 + 2

b) dom: [1, infty), ran: [2, infty)

Question 2

a) f(g(x)) = rt[x+(1/x)-1]

b) g(f(x)) = rt[x-1] + 1/rt[x-1]

Question 3

y = 2rt[x+1] - 2

Question 4

The gradients are -3/4 and 2 respectively, hence the angle made with horizontal is tan-1(-3/4) and tan-1(2), which is taking the difference between them, you get an angle of ~100 degrees, hence the acute angle between them would be ~80 degrees, tan(80.xxxx) ~= 5.5 (keeping the decimals)

There is an analytical way to do this, but that will involve the compound angle for Tan, which is not part of Mathematical Methods (CAS).

Question 5

I'm pretty sure you have mis-typed that question. There is no constant 'c' term. The factorised form of what you have given (2x2 - x - x) is 2x(x-1)

May I suggest www.vcenotes.com as an alternate source for asking mathematical questions, since the forum there has LaTeX capabilities, and answers are properly displayed like you see in a textbook. We also have many more mathematically enthusiastic students/tutors who usually respond to questions very fast (at most within a few hours).

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Actually, if you need help with maths, and you liked Merspi, mathoverflow.net is a far better option. – Collin Li♦ Dec 23 at 14:39
Fixed your formatting! – Collin Li♦ Dec 23 at 14:48
would you be able to explain your answer to question 1,2 and 3? I don't understand the working out fully (there isn't any) thanks for the very commited answer! – Impractical Dec 23 at 15:46
I have to go soon, but I'll type out Question 1 :) – Collin Li♦ Dec 23 at 22:14
1 
 Re: LaTeX, we believe this is not a problem (and believe it is feature-creep), as we can use <sup> and <sub> HTML tabs, and relatively easy formatting here. The bigger problem that is yet to be solved by anyone is easily embedded pictures. That will solve both the typeset problem and the graphical problem. – Collin Li♦ Dec 24 at 7:09

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