Solve Inequality Laws for x in Spivak's Calculus

[Gamerex]

I just got Spivak's calculus today, and I'm already stuck on the prologue problems:

1. The problem
Find all x for which (x-1)(x-3)>0


2. The attempt at a solution

We know that if ab>0, then either a>0 and b>0, or a<0 and b<0.

Thus, if a=(x-1) and b=(x-3), then either (x-1)>0 and (x-3)>0, or (x-1)<0 and (x-3)<0.

Solving for x would yield four solutions, but only two, x<1 and x>3, are true. Why is this?

 

[scurty]

You need to consider the values of above, below, and in between all of the points that make the equation equal to zero. In this case there are three ranges you need to check. You need to check these ranges because the sign value can change from range to range.

 

[Gamerex]

But I don't understand how the definition, "if ab>0, then either a>0 and b>0, or a<0 and b<0" can remain true in this instance. In other words, I know what to do, but I don't know why it works.

 

[scurty]

Because a and b are both "functions" of x. In your case they are just numbers, changing one number doesn't affect the other. Does that make sense?

 

[Gamerex]

If I'm correct, you're saying "if ab>0, then either a>0 and b>0, or a<0 and b<0"
is only true if a and b are functions.

But I considered "a=(x-1) and b=(x-3)". Aren't x-1 and x-3 functions of x?

 

[Gamerex]

Oh wait, the solution just hit me!
For "if ab>0, then either a>0 and b>0, or a<0 and b<0" to be true,
a and b have to ALWAYS be either greater or less than 0. In other words, a and b can never change signs.

Obviously, (x-1) and (x-3) DO change signs at x=1 and x=3 respectively, so the conditions for that definition are not met; thus, it does not apply.

Thanks for your help, I would have never gotten it if you hadn't mentioned functions.

 

[HallsofIvy]

I just got Spivak's calculus today, and I'm already stuck on the prologue problems:

1. The problem
Find all x for which (x-1)(x-3)>0


2. The attempt at a solution

We know that if ab>0, then either a>0 and b>0, or a<0 and b<0.

Thus, if a=(x-1) and b=(x-3), then either (x-1)>0 and (x-3)>0, or (x-1)<0 and (x-3)<0.

Solving for x would yield four solutions, but only two, x<1 and x>3, are true. Why is this?

x- 1> 0 and x- 3>0 give x> 1 and x> 3. In order that both of these be true, we must have x> 3. (x= 2 satisfies x> 1 but not x> 3 so does not satisfy (x- 1)(x- 3)> 0. x> 1 alone is not enough.

x- 1< 0 and x- 3< 0 give x< 1 and x< 3. In order that both of these be true, we must have x< 1.

 

[azizlwl]

(x-1)(x-3)>0

I prefer to use set operations(if you have studied), union and intersection to see more clearly.

[itex]((x-1)>0 \cap (x-3)>0) \ \ \ \bigcup \ \ \ ((x-1)<0 \cap (x-3)<0)[/itex]
[itex]x>1 \cap x>3 \ \ \bigcup \ \ x<1\ \cap x<3[/itex]

As you see on the left of bigCup, x>3 also means >1. 2 even >1 but <3
On the right side, anything less than 1 is also less than 3

So have x>3 or x<1
x=(-∞,1) or (3,+∞)