# Solve Inequality Laws for x in Spivak's Calculus

• Gamerex
In summary, to solve the equation (x-1)(x-3)>0, we must consider the ranges of x where both a=(x-1) and b=(x-3) are either both greater or both less than 0. This leads to the solutions x>3 or x<1. In other words, the solution set is x=(-∞,1) or (3,+∞). The definition "if ab>0, then either a>0 and b>0, or a<0 and b<0" does not apply in this instance because a and b are not always greater or less than 0, as x=1 and x=3 make them equal to 0.
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?

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.

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.

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?

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?

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.

Gamerex said:
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.

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

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

$((x-1)>0 \cap (x-3)>0) \ \ \ \bigcup \ \ \ ((x-1)<0 \cap (x-3)<0)$
$x>1 \cap x>3 \ \ \bigcup \ \ x<1\ \cap x<3$

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,+∞)

## 1. How do I solve inequality laws for x in Spivak's Calculus?

To solve inequality laws for x in Spivak's Calculus, you can follow these steps:

• 1. Identify the given inequality and determine the appropriate inequality symbol (greater than, less than, greater than or equal to, or less than or equal to).
• 2. Isolate the variable x on one side of the inequality by using algebraic operations (addition, subtraction, multiplication, division) to get rid of any constants or other variables.
• 3. If you multiply or divide by a negative number, remember to flip the inequality symbol to maintain the same inequality.
• 4. Graph the solution on a number line or use interval notation to express the solution.

## 2. What are the most common inequality symbols used in Spivak's Calculus?

The most common inequality symbols used in Spivak's Calculus are: <, >, ≤, and ≥. These symbols represent less than, greater than, less than or equal to, and greater than or equal to, respectively.

## 3. Can you solve inequality laws for x in Spivak's Calculus without using algebra?

No, in order to solve inequality laws for x in Spivak's Calculus, you need to use algebraic operations and properties to isolate the variable x and find its value. Without using algebra, it would be difficult to manipulate the given inequality and find the solution.

## 4. Are there any special cases when solving inequality laws for x in Spivak's Calculus?

Yes, there are a few special cases to keep in mind when solving inequality laws for x in Spivak's Calculus. These include:

• - If the inequality symbol is a < or >, the solution will be an open interval.
• - If the inequality symbol is a ≤ or ≥, the solution will be a closed interval.
• - If the inequality symbol is a ≤ or ≥ and there is a negative number involved, the solution will be a reversed interval.

## 5. How can I check my answer when solving inequality laws for x in Spivak's Calculus?

To check your answer when solving inequality laws for x in Spivak's Calculus, you can substitute your solution back into the original inequality and see if it makes the inequality true. You can also graph the solution on a number line and visually check if it satisfies the given inequality.

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