Question

NO LINKS!! Please help me with this problem Part 4gg​

Answer 1

(-1, 1) ∪ (3, ∞)

Explanation:

Given inequality:

`x^3-3x^2-x+3 > 0`

Solve the inequality as though it were an equation.

`\implies x^3-3x^2-x+3=0`

Factor the equation.

If the coefficients in a polynomial add up to 0, then (x - 1) is a factor.

`\implies (x-1)(x^2-2x-3)=0`

`\implies(x-1)(x+1)(x-3)=0`

`\implies x-1=0 \implies x=1`

`\implies x+1=0 \implies x=-1`

`\implies x-3=0 \implies x=3`

The real solutions to the equation are the boundary points for the solution to the inequality.

Make the boundary points open circles as the original inequality does not include equality.

Four regions have been created:

• x < -1
• -1 < x < 1
• 1 < x < 3
• x > 3

Select points from the different regions and test to see if they satisfy the original inequality:

`x=-2 \implies (-2)^3-3(-2)^2-(-2)+3=-15 < 0`

`x=0 \implies (0)^3-3(0)^2-(0)+3=3 > 0`

`x=2 \implies (2)^3-3(2)^2-(2)+3=-3 < 0`

`x=4 \implies (4)^3-3(4)^2-(4)+3=15 > 0`

Since x = -2 does not satisfy the original inequality, the region x < -1 is not part of the solution.

Since x = 0 does satisfy the original inequality, the region -1 < x < 1 is part of the solution.

Since x = 2 does not satisfy the original inequality, the region 1 < x < 3 is not part of the solution.

Since x = 4 does satisfy the original inequality, the region x > 3 is part of the solution.

Therefore, the solution in interval notation is:

• (-1, 1) ∪ (3, ∞)

To graph the solution set:

• Place open circles at x = -1, x = 1 and x = 3.
• Connect the open circles at x = -1 and x = 1 with a line.
• Draw a line to the right of the open circle at x = 3 and place an arrow at the other end of the line to indicate it continues indefinitely.