Solve the inequality and put in interval notation: 2x^3-2x≥0

Answer 1

To solve polynomial inequalities, you want to start by treating them like equations. Set it "equal" to zero and "solve". This one is already set equal to zero, so we just solve. The first step there is to factor out a 2x to leave you with:

2x(x^2 -1) >= 0

Now you can set each factor equal to zero:

2x = 0 x^2 - 1 = 0

Solve each of those equations:

x = 0 x^2 = 1

x = +/- 1

Remember, we want to solve 2x(x^2 -1) >= 0. We've just found three values (-1, 0, 1) that solve the equation 2x(x^2 -1) = 0. For inequalities, you want to think about a number line. Any point on the number, the left side of your inequality is either positive, negative, or 0. We're looking for >=0 so we want the values where it's positive or 0. We already know where it equals 0, and those are also the only points where it will change from positive to negative. So essentially, we've just broken the number line up at three points:

___________________________|______________|________________|___________________________

-1 0 1

Pick a test point in each of those intervals, a number less than -1, a number between -1 and 0, a number between 0 and 1, and a number bigger than 1. Plug the test points into the left side of the inequality to see whether the value is positive or negative. I'll use -2, -1/2, 1/2, and 2:

2(-2) ((-2)^2 - 1) = -4(3) = -12...we're negative to the left of -1

2(-1/2) ((-1/2)^2 - 1) = -1(-3/4) = 3/4...we're positive between -1 and 0

2(1/2) ((1/2)^2 - 1) = 1(-3/4) = -3/4...we're negative between 0 and 1

2(2) ((2)^2 - 1) = 4(3) = 12...we're positive to the right of 1

So again, our solutions are any values where the left side was equal to zero or positive. For interval notation, your intervals give beginning and ending values where you have solutions. You use brackets if the beginning or ending value is a solution (if it's included).

We have solutions between -1 and 0 (including AT -1 and 0), and anything bigger than 1 (including AT one). That means we need two different intervals and they're joined by a union symbol:

[-1, 0] U [1, inf)

Answer 2

2x³ - 2x ≥ 0

2x(x² - 1) ≥ 0 <-------Factor as much as possible

2x(x + 1)(x - 1) ≥ 0

Notice that when x = ±1, 0, this is where the inequality is zero. So these values are part of our answer for sure. Now you make what's called a sign chart.

-1 0 1 <-------------- You'll notice that the zeros are in ascending order

2x(x+1)(x-1) - + - + <------------All you're doing here is figuring out where numbers are positive or negative.

So here's how it works: If you choose a number smaller than -1 (let's say you choose x = -2) Then when you plug in x = -2 into the expression 2x(x + 1)(x - 1) = 2(-2)(-2 + 1)(-2 -1) is a NEGATIVE number. So all numbers lower than -1 will be negative. Now choose a number between -1 and 0, like -0.5. Plug this into 2x(x + 1)(x - 1) = 2(-0.5)(-0.5 + 1)(-0.5 - 1) is POSITIVE. Thus, any number you pick between -1 and 0 will be positive. And you keep filling this chart out as shown above.

Now, since your problem says that you want to solve the inequality:

2x(x + 1)(x - 1) ≥ 0

Then you are looking for x - values that make the inequality POSITIVE. (Since the inequality is saying greater than zero).

Thus, when you look at your sign chart, your answer would be:

[-1, 0] ∪ [1, ∞).

Hope this helps.

Mr. K

(Sorry if the sign chart is coming out funny, I tried editing it 5 times now.)

Answer 3

2x³ - 2x ≥ 0

2x(x² - 1) ≥ 0

2x(x + 1)(x - 1) ≥ 0

For the equality x = -1, x = 0, and x = 1

For the greater than all factors must be positive, or two of the factors negative and the other positive.

For all to be positive x > 1

For two negative and one positive -1 < x < 0

[-1, 0] U [1, ∞)