solve the following inequality. write your solutionin interval notation x^3-x<0
x(x-1)(x+1) < 0
The roots are -1,0,1
Since there are no repeated roots, the graph crosses the axis at each root.
So, y<0 for large x<0, so the intervals are
(oo,-1) < 0
(-1,0) > 0
(0,1) < 0
(1,oo) > 0
y<0 for x in (-oo,-1)U(0,1)
To solve the inequality x^3 - x < 0, we need to find the values of x that make the inequality true. Here's how you can do it step by step:
Step 1: Factor the inequality if possible.
Since x^3 - x is a polynomial expression and can't be easily factored, we need to use a different approach.
Step 2: Determine the critical points.
To find the critical points, we set x^3 - x equal to 0 and solve for x. By factoring out an x, we get:
x(x^2 - 1) = 0
Now, we have two factors: x = 0 and x^2 - 1 = 0. Solving the quadratic equation x^2 - 1 = 0, we have:
x^2 = 1
x = ±1
So the critical points are x = 0, x = -1, and x = 1.
Step 3: Create a sign chart.
To create a sign chart, we divide the number line into four intervals: (-∞, -1), (-1, 0), (0, 1), and (1, ∞). We also include the critical points on the chart.
Step 4: Test a value in each interval.
To determine the sign of x^3 - x in each interval, we can choose a test value from each interval and substitute it into the expression. We don't need to find the exact values; we just need to know if the result is positive or negative.
Let's choose x = -2, -0.5, 0.5, and 2 as our test values.
For the interval (-∞, -1), we substitute x = -2:
(-2)^3 - (-2) = -8 + 2 = -6 (negative)
For the interval (-1, 0), we substitute x = -0.5:
(-0.5)^3 - (-0.5) = -0.125 + 0.5 = 0.375 (positive)
For the interval (0, 1), we substitute x = 0.5:
(0.5)^3 - (0.5) = 0.125 - 0.5 = -0.375 (negative)
For the interval (1, ∞), we substitute x = 2:
(2)^3 - (2) = 8 - 2 = 6 (positive)
Step 5: Analyze the sign chart and determine the solution.
Looking at the sign chart, we can see that the expression x^3 - x is negative (less than 0) for the intervals (-∞, -1) and (0, 1). It is positive (greater than 0) for the interval (-1, 0) and (1, ∞).
Therefore, our solution in interval notation is:
(-∞, -1) U (0, 1)