Understand the Properties of Functions

mvola
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Homework Statement



Given f(x) = 1-x
Show that 2f(3x + 1) = -6x

Homework Equations





The Attempt at a Solution


I've tried many approaches, but I seem to be missing something. The latest approach was as follows and did not work.

Divide by 2.
f(3x + 1) = -3x
f(3x) + f(1) = -3x
3f(x) + f(1) = -3x
3(1-x) + 0 = -3x
3 - 3x = -3x
3 = 0 which is obviously not true.

Can anyone help me out with this?
 
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mvola said:

Homework Statement



Given f(x) = 1-x
Show that 2f(3x + 1) = -6x

Homework Equations





The Attempt at a Solution


I've tried many approaches, but I seem to be missing something. The latest approach was as follows and did not work.

Divide by 2.
f(3x + 1) = -3x
f(3x) + f(1) = -3x
3f(x) + f(1) = -3x
3(1-x) + 0 = -3x
3 - 3x = -3x
3 = 0 which is obviously not true.

Can anyone help me out with this?

You are going about this completely the wrong way and using properties of f that aren't even true. To find f(3x+1) just substitute (3x+1) for x in 1-x. Doesn't that make more sense?
 
mvola said:

Homework Statement



Given f(x) = 1-x
Show that 2f(3x + 1) = -6x

##2f(3x+1)=2(1-(3x+1))=2(-3x)=-6x##.

Think of it this way: If ##f(x)=1-x##, then ##f(Lemon \, cake) = 1 - (Lemon \, cake)##, in the unlikely event that the quantity 1-(Lemon cake) is defined.
 
I was over-thinking it. Thank you!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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