Question

Find the points on the curve \( y = 2x^3 - 3x^2 - 12x + 15 \) where rate of change of dependent variable with respect to the independent variable vanishes. What do we call these type of points?

Hint:

Vanishing derivative indicates that the tangent to the curve is horizontal at those points. These are locations where local maxima or minima might occur.

Answer 1

Step 1: Understanding the Concept:
The rate of change of the dependent variable \( y \) with respect to the independent variable \( x \) is the derivative \( \frac{dy}{dx} \). When this "vanishes," it means \( \frac{dy}{dx} = 0 \).
Step 2: Key Formula or Approach:
Find \( \frac{dy}{dx} \), set it to zero, and solve for \( x \). Then find the corresponding \( y \) values.
Step 3: Detailed Explanation:
Given \( y = 2x^3 - 3x^2 - 12x + 15 \).
Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = 6x^2 - 6x - 12 \] Set \( \frac{dy}{dx} = 0 \):
\[ 6(x^2 - x - 2) = 0 \] \[ x^2 - x - 2 = 0 \] Factoring the quadratic:
\[ (x-2)(x+1) = 0 \implies x = 2, x = -1 \] Now find the corresponding \( y \) values:
For \( x = 2 \):
\[ y = 2(2)^3 - 3(2)^2 - 12(2) + 15 = 16 - 12 - 24 + 15 = -5 \] So, point 1 is \( (2, -5) \).
For \( x = -1 \):
\[ y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 15 = -2 - 3 + 12 + 15 = 22 \] So, point 2 is \( (-1, 22) \).
Points where the first derivative is zero are called Stationary Points or Critical Points.
Step 4: Final Answer:
The points are \( (2, -5) \) and \( (-1, 22) \). These are called stationary points.