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Solve each inequality. Use the number line provided to test intervals.

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Solve each inequality. Use the number line provided to test intervals. Thank you-example-1

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Answer:

[-2, 2/5] ∪ [2, ∞)

Explanation:

You want the solution to the inequality 7x³ -20x +11 ≥ 2x³ +2x² +3.

Factors

The inequality can be rewritten to a comparison with zero by subtracting the right-side expression.

7x³ -20x +11 -(2x³ +2x² +3) ≥ 0

5x³ -2x² -20x +8 ≥ 0 . . . . . . . simplify

x²(5x -2) -4(5x -2) ≥ 0 . . . . . . factor pairs of terms

(x² -4)(5x -2) ≥ 0 . . . . . . . . . . remove common factor

(x -2)(5x -2)(x +2) ≥ 0 . . . . . factor difference of squares

Intervals

The values of x that make this product equal to zero are

x -2 = 0 ⇒ x = 2

5x -2 = 0 ⇒ x = 2/5

x +2 = 0 ⇒ x = -2

These x-intercepts divide the domain into 4 intervals where we need to check the sign.

From left to right:

(-∞, -2)

Using x = -3, we find the product to be ...

(-3 -2)(5(-3) -2)(-3 +2) = (-5)(-17)(-1) = -85, less than 0

[-2, 2/5]

Using x = 0, we find the product to be ...

(0 -2)(5·0 -2)(0 +2) = (-2)(-2)(2) = 8, part of the solution

(2/5, 2)

Using x = 1, we find the product to be ...

(1 -2)(5·1 -2)(1 +2) = (-1)(3)(3) = -9, less than 0

[2, ∞)

Using x = 3, we find the product to be ...

(3 -2)(5·3 -2)(3 +2) = (1)(13)(5) = 65, part of the solution

The solution to the inequality is [-2, 2/5] ∪ [2, ∞).

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Additional comment

The function we're comparing to zero is a cubic with a positive leading coefficient. That means the general shape of its graph is /, negative on the left and positive on the right.

The graph changes sign at each of the zeros, so we already know the curve is positive between the left two x-intercepts, and to the right of the rightmost x-intercept. We just need to define the intervals so the zeros themselves are included in the solution set.

Solve each inequality. Use the number line provided to test intervals. Thank you-example-1
Solve each inequality. Use the number line provided to test intervals. Thank you-example-2
User Asif Iqbal
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