Question

Graph the function y = 2x3 – x2 – 4x + 5. To the nearest tenth, over which interval is the function decreasing?

(1, ∞)
(–∞, –0.7)
(–0.7, 1)
(–1, 0.7)

Answer 1

To determine the intervals where the function \( y = 2x^3 - x^2 - 4x + 5 \) is decreasing, we need to find the derivative of the function and analyze its critical points.

1. **Find the derivative**:
\[
y' = \frac{dy}{dx} = 6x^2 - 2x - 4
\]

2. **Set the derivative equal to zero to find critical points**:
\[
6x^2 - 2x - 4 = 0
\]
This simplifies to:
\[
3x^2 - x - 2 = 0
\]

3. **Use the quadratic formula to solve for \( x \)**:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3}
\]
\[
= \frac{1 \pm \sqrt{1 + 24}}{6} = \frac{1 \pm \sqrt{25}}{6} = \frac{1 \pm 5}{6}
\]
The critical points are:
\[
x = \frac{6}{6} = 1 \quad \text{and} \quad x = \frac{-4}{6} = -\frac{2}{3} \approx -0.7
\]

4. **Determine the sign of the derivative in each interval**:
We need to check the intervals created by the critical points: \( (-\infty, -\frac{2}{3}) \), \( (-\frac{2}{3}, 1) \), and \( (1, \infty) \).

- For \( x < -\frac{2}{3} \) (choose \( x = -1 \)):
\[
y'(-1) = 6(-1)^2 - 2(-1) - 4 = 6 + 2 - 4 = 4 \quad \text{(positive)}
\]

- For \( -\frac{2}{3} < x < 1 \) (choose \( x = 0 \)):
\[
y'(0) = 6(0)^2 - 2(0) - 4 = -4 \quad \text{(negative)}
\]

- For \( x > 1 \) (choose \( x = 2 \)):
\[
y'(2) = 6(2)^2 - 2(2) - 4 = 24 - 4 - 4 = 16 \quad \text{(positive)}
\]

5. **Summary of intervals**:
- \( (-\infty, -\frac{2}{3}) \): \( y' > 0 \) (increasing)
- \( (-\frac{2}{3}, 1) \): \( y' < 0 \) (decreasing)
- \( (1, \infty) \): \( y' > 0 \) (increasing)

Thus, the function is decreasing on the interval \((-0.7, 1)\).

The answer is:
\[
(-0.7, 1)
\]