Basic Calculus Inequlities Questions?

Question

1) x^3 + 4x < 0

2) 1/(2x+1) < x/(x-3)

Answer 1

(1).

The inequality is x³ + 4x < 0.

Write the polynomial as factor form

x³ + 4x = x(x² + 4).

The key numbers are x = 0 and x = ± 2i. Here the imaginary numbers are negligible. So, the polynomial’s test intervals are (-∞, 0) and (0, ∞).

In each test interval, choose a representative x-value and evaluate the polynomial.

Test Interval  x-value                Polynomial Value                        Conclusion

(-∞, 0)            x = - 1              (- 1)³ + 4(- 1) = - 1 - 5 = - 6 < 0            Negative

(0, ∞)              x = 1                (1)³ + 4(1) = 1 + 4 = 5 < 0                    Positive

From this you can conclude that the inequality is satisfied on the open intervals (-∞, 0). So, the solution set is (-∞, 0) and its graph is

Answer 2

(2).

The rational inequality is 1/(2x + 1) < x/(x - 3).

write the inequality in general form with the rational expression on the left and zero on the right.

Subtract x/(x - 3) from each side.

[1 / (2x + 1)] - [x / (x - 3)] < 0

[(x - 3) - x(2x + 1)] / [(2x + 1)(x - 3)] < 0

[x - 3 - 2x² - x] / [(2x + 1)(x - 3)] < 0

 - (3 + 2x²) / [(2x + 1)(x - 3)] < 0

Multiply each side by negative 1 and reverse the inequality symbol.

(3 + 2x²) / [(2x + 1)(x - 3)] > 0

The zeros (the x-values for which its numerator is zero) and its undefined values (the x-values for which its denominator is zero) of the rational expression are called key numbers.

Numerator is zero, 3 + 2x² = 0 ⇒ x = ± i√(3/2).

Denominator is zero, (2x + 1)(x - 3) = 0 ⇒ x = - 1/2 and x = 3.

The key numbers are x = ± i√(3/2), x = - 1/2 and x = 3. Here the imaginary numbers are negligible. So, the polynomial’s test intervals are (- ∞, - 1/2), (- 1/2, 3) and (3, ∞).

In each test interval, choose a representative x-value and evaluate the polynomial.

Test Interval  x-value  Polynomial Value [(x + 11)/(x + 4) < 0]             Conclusion

(-∞, -1/2)          x = - 1   [3 + 2(-1)²] / [{2(-1) + 1} {(-1) - 3}] = 5/4 > 0           Positive

(- 1/2, 3)          x = 0    [3 + 2(0)²] / [{2(0) + 1} {(0) - 3}] = 3/(-3) = - 1 < 0   Negative

(3, ∞)               x = 4      [3 + 2(4)²] / [{2(4) + 1} {(4) - 3}] = 35/9 > 0             Positive

From this we can conclude that the inequality is satisfied on the open intervals (-∞, -1/2) and (3, ∞). So, the solution set is (-∞, -1/2) U (3, ∞) and its graph is