Function problems: p(x) = 3x + a and p^-1(x)=3bx - 4/3

[RonHKA]

Hey all,


I've been trying to solve these two questions concerning functions for quite some time and I'm running out of solutions. Anyway, here are the questions.

a) Given p(x) = 3x + a and p^-1(x)=3bx - 4/3 , where a ad b are constants. Find the value of a and b. (y-

What I currently have:
p(x) = 3x + a
let y = p(x)
y = 3x+a
(y-a)/3 = x
p^1(x) = (x-a)/3
(x-a)/3 = 3bx - 4/3
At this point, I have no idea what to do to get values a and b. :/

b) Given functions f^-1(x) = (3-mx)/2 and g(x)= 2x^2 - 3. Find the value of m such that g(-x) = 1/4 f(x^2)

What I currently have:
y = (3-mx)/2
2y = 3-mx
mx = 3 - 2y
x = (3-2y)/m
f(x) = (3-2x)/m

g(-x) = 1/4 ((3-2x^2)/m)
2x - 3 = 1/4 ((3-2x^2)/m)


At this point, I am lost. lol :/






I find these question hard and no matter what I do, I can't seem to get the right answer. Any help is appreciated. Thanks in advanced.

 

[Ishuda]

Let's do a: You got to the point where you had
(x-a)/3 = 3bx - 4/3
or, multiplying both sides by 3
x - a = 9 b x - 4.
Now that has to hold for all x, so let's put in x=0 and we have -a=-4 or a = 4 and
p(x) = 3x + 4
and leads to
x - 4 = 9 b x - 4
Now do the same sort of thing to find b (maybe x=1 or even x = 1/9).

BTW, it is always a good idea to test your answer if you have time.

For (b), you did f(x) correctly by made a mistake following that:
g(-x) = 1/4 ((3-2x^2)/m)
2x^2 - 3 = 1/4 ((3-2x^2)/m)

At this point, I would suggest you multiply through by 4 m and then do the same kind of thing as suggested for (a). Note that the 2x² for g(-x) is correct only because (-x)² = x².