Solve the inequality.
f(x)> 0, where f(x) =-x(x+1)(x-1)(x-3)
To solve the inequality f(x) > 0, where f(x) = -x(x+1)(x-1)(x-3), we need to find the intervals of x values that satisfy the inequality.
First, we find the critical points of the function by setting f(x) equal to zero and solving for x:
- x(x+1)(x-1)(x-3) = 0
Now, we have four critical points: x = -1, x = 0, x = 1, and x = 3.
We can use these critical points to divide the number line into five regions: x < -1, -1 < x < 0, 0 < x < 1, 1 < x < 3, and x > 3.
To determine whether f(x) is positive or negative in each of these regions, we can choose a test point within each region and evaluate f(x) at that point.
For the region x < -1, we can choose x = -2. Plugging it into f(x), we get:
f(-2) = -(-2)(-2+1)(-2-1)(-2-3) = -2(1)(-3)(-5) = 30
Since f(-2) is positive, f(x) is positive in the region x < -1.
For the region -1 < x < 0, we can choose x = -0.5. Evaluating f(x) at this point:
f(-0.5) = -(-0.5)(-0.5+1)(-0.5-1)(-0.5-3) = 0.5(0.5)(-1.5)(-3.5) = 3.9375
Since f(-0.5) is positive, f(x) is positive in the region -1 < x < 0.
For the region 0 < x < 1, we can choose x = 0.5. Evaluating f(x) at this point:
f(0.5) = -(0.5)(0.5+1)(0.5-1)(0.5-3) = -0.5(1.5)(-0.5)(-2.5) = 0.9375
Since f(0.5) is positive, f(x) is positive in the region 0 < x < 1.
For the region 1 < x < 3, we can choose x = 2. Evaluating f(x) at this point:
f(2) = -(2)(2+1)(2-1)(2-3) = -2(3)(1)(-1) = 6
Since f(2) is positive, f(x) is positive in the region 1 < x < 3.
Finally, for x > 3, we can choose x = 4. Evaluating f(x) at this point:
f(4) = -(4)(4+1)(4-1)(4-3) = -4(5)(3)(1) = -60
Since f(4) is negative, f(x) is negative in the region x > 3.
Therefore, the solution to the inequality f(x) > 0 is:
x < -1 or -1 < x < 0 or 0 < x < 1 or 1 < x < 3