Show that the function y = 2x3+3x2−12x+1 decreases in the interval (−2,1).

To show that the function \( y = 2x^3 + 3x^2 - 12x + 1 \) decreases in the interval \( (-2,1) \), we need to follow these steps:

### Step 1: Compute the first derivative of the function

The first derivative \( y'(x) \) of the function represents the rate of change of \( y \) with respect to \( x \). If \( y'(x) < 0 \) in an interval, the function is decreasing in that interval.

Given:
\[
y(x) = 2x^3 + 3x^2 - 12x + 1
\]
Differentiate \( y(x) \) with respect to \( x \):
\[
y'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 12x + 1)
\]
\[
y'(x) = 6x^2 + 6x - 12
\]

### Step 2: Analyze the sign of the first derivative in the interval \( (-2,1) \)

We now need to determine if \( y'(x) \) is negative in the interval \( (-2, 1) \). This involves finding the critical points by solving \( y'(x) = 0 \) and testing the sign of \( y'(x) \) within the interval.

Solve \( 6x^2 + 6x - 12 = 0 \) for \( x \):
\[
x^2 + x - 2 = 0
\]
Factor the quadratic:
\[
(x + 2)(x - 1) = 0
\]
Thus, the critical points are \( x = -2 \) and \( x = 1 \).

### Step 3: Sign test in the interval \( (-2,1) \)

We will now test the sign of \( y'(x) \) in the interval \( (-2,1) \). Pick test points in the subintervals:
- For \( x = -1 \) (in \( (-2,1) \)):
\[
y'(-1) = 6(-1)^2 + 6(-1) - 12 = 6 - 6 - 12 = -12
\]
Thus, \( y'(-1) < 0 \).

Since \( y'(x) = 6x^2 + 6x - 12 \) is a continuous function, and it is negative at \( x = -1 \) and the critical points at \( x = -2 \) and \( x = 1 \) mark the boundaries, we conclude that \( y'(x) < 0 \) in the entire interval \( (-2,1) \).

### Conclusion

The function \( y = 2x^3 + 3x^2 - 12x + 1 \) decreases in the interval \( (-2, 1) \) because its derivative \( y'(x) \) is negative throughout this interval.

Would you like more details on any step, or do you have any questions?

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Here are 5 related questions you might find useful:

1. How do you find intervals of increase and decrease for a function?
2. What is the significance of the first derivative test?
3. How do you analyze critical points in more complex functions?
4. How do second derivatives help in determining concavity?
5. How would you apply this method to trigonometric functions?

**Tip**: To confirm whether a function is increasing or decreasing in an interval, always check the sign of the derivative.

Show that the function y = 2x^3 + 3x^2 − 12x + 1 decreases in the interval (−2,1).

Learn how to show that the function y = 2x^3 + 3x^2 - 12x + 1 decreases in the interval (−2,1) using the first derivative test. This step-by-step guide covers calculus concepts like finding the derivative, solving critical points, and testing the sign of the derivative. Suitable for high school calculus students.

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