Need help with math.

[mittacc]

Hello!

Right now I'm doing some math since I don't have anything better to do atm and it is always good to be ahead of schedule. I've come across a problem which I cannot solve by myself and I would like some help.

The problem is following:

Let f(x) = ax + b. Define a and b so f(ax + b) = x + 1.

The answer sheet says that gives a squared = 1 and that gives b = 0.5.

I can't comprehend this, any help would be appreciated!

[belfpala]

f(x) = ax + b

Then f(ax + b) = a(ax + b) + b = x + 1, right?

Mess around from there, the answer given appears correct.

[mittacc]

I've gotten to a^2 x + ab + b= x +1 / divide by x
a^2 + ab + b = 1

Which proves that either I or the answer sheet is incorrect(since a squared should be = 1), dunno which though :/

From there i could go to
a=(1/a) - 2b but I don't see the point in that :/

[belfpala]

That divide by x isn't going to work. you'd end up with a^2 + ab/x + b/x = 1 + 1/x

[stalkerzzzz]

a^2 x + ab + b= x +1 => a^2 = 1 and that ab + b = 1
a^2 = 1 => a=+1,-1
a=1 => b+b=1 => b=0.5
a=-1 => -b+b=1 => 0=1 false => b can only be 0.5

[heheMongot]

f(x)=ax + b
f(ax+b)= a(ax+b)+b = a^2*x + ab + b

Since the only factor containing an x is (a^2*x) that must mean that the coefficient (a^2) must be equal to the coefficient of (x) which is then 1.
From here you see that the remaining factors (ab + b) need to make up the constant in the right side of your equation (1).
Either you just see from here that in order for (ab + b) to be 1 (b) needs to be (0,5) since (a=1) or you could solve it like this:

ab+b = 1
a=1
1b + b = 1
2b = 1
b = 1/2

This leaves us with a=1 and b=0,5

Hope this helps.

[mittacc]

f(x)=ax + b
f(ax+b)= a(ax+b)+b = a^2*x + ab + b

Since the only factor containing an x is (a^2*x) that must mean that the coefficient (a^2) must be equal to the coefficient of (x) which is then 1.
From here you see that the remaining factors (ab + b) need to make up the constant in the right side of your equation (1).
Either you just see from here that in order for (ab + b) to be 1 (b) needs to be (0,5) since (a=1) or you could solve it like this:

ab+b = 1
a=1
1b + b = 1
2b = 1
b = 1/2

This leaves us with a=1 and b=0,5

Hope this helps.

Cleared it up quite a bit, but I bolded the part which confused me. How am I supposed to know that a squared * x = x and not 1?

[heheMongot]

Then b has to be x which it could. Another solution would be a=0 and b=x+1.

[Elhana]

General solution:
a(ax + b) + b = x + 1 =>
(a^2-1)x = -(a+1)b+1
Now to find solution we divide both parts by (a^2-1), but we have to check a case where it is 0:
1) a^2-1 = 0 => a^2 = 1, again 2 cases:
1.1) a=1 => 0 = -2b+1 => b = 1/2 (our answer)
1.2) a=-1 => 0 = 1 - false

2) a^2 != 1 ; x = -((a+1)b+1)/(a^2-1) - doesn't fit original task since any given a/b supposed to be constant (I assume) and independent from x, but technically heheMongot is somewhat correct, since a != -1; b=(1+x-a^2x)/(a+1) is also an answer.

But in your case you should realise that in a function f(x) = ax+b, coefficient a defines an angle (it's tangens to be precise) between your line and x-axis and b defines it's shift along the axis.
For two line functions to overlap, their a and b coefficiets have to be the same, otherwise they will only cross in one point or never. So we split original equation in 2 parts:
a^2 (x) = 1 (x) & ab+b = 1 =>
a = 1, b = 1/2

[Cairhiin]

f(x) = ax + b

f(ax + b) = a(ax + b) + b = a^2(x) + ab + b

Since y = x + 1 is a linear equation with slope of +1 and a y-intercept of +1 f(ax + b) needs to have the same slope and same y-intercept.

The slope coefficient of f(ax + b) is a^2 and needs to be equal to 1. a^2 = 1 a = +sqrt(1) or -sqrt(1) (a = -1, +1)

To find the y-intercept (b) we solve ab + b = 1 (this is the static portion of the linear equation and thus corresponds to the y-intercept) for both a = -1 and a = 1.

a = 1: 1b + 1b = 1 => b = 1/2
a = -1: -1b + b = 1 => 0 <> 1 (no solution)

Thus for both lines to be exactly the same a needs to be 1, and b needs to be 1/2.