Question

The curve \( 4y = 3x^4 - 2x^2 \) attains

• A. both minimum values
• B. a maximum value and a minimum value respectively
• C. a minimum value and a maximum value respectively
• D. both maximum values

Hint:

In curve analysis, the second derivative test helps determine whether a critical point corresponds to a maximum or minimum. If \( \frac{d^2y}{dx^2} > 0 \), the point is a minimum, and if \( \frac{d^2y}{dx^2} < 0 \), the point is a maximum.

Answer 1

The Correct Option is A

Step 1: Equation of the curve.
The given curve is \( 4y = 3x^4 - 2x^2 \).

Step 2: First derivative of the curve.
We first find the first derivative \( \frac{dy}{dx} \). Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(4y) = \frac{d}{dx}(3x^4 - 2x^2) \] \[ 4 \frac{dy}{dx} = 12x^3 - 4x \] Thus, the first derivative is:
\[ \frac{dy}{dx} = 3x^3 - x \]

Step 3: Find critical points.
To find the critical points, we set \( \frac{dy}{dx} = 0 \):
\[ 3x^3 - x = 0 \] \[ x(3x^2 - 1) = 0 \] So, \( x = 0 \) or \( x = \pm \frac{1}{\sqrt{3}} \).

Step 4: Second derivative test.
Next, we find the second derivative to check if the critical points correspond to maxima or minima.
\[ \frac{d^2y}{dx^2} = 9x^2 - 1 \]

Step 5: Evaluate at \( x = -\frac{1}{\sqrt{3}} \) and \( x = \frac{1}{\sqrt{3}} \).
- At \( x = -\frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} = 9 \left( -\frac{1}{\sqrt{3}} \right)^2 - 1 = 3 - 1 = 2 \), which is positive, indicating a minimum.
- At \( x = \frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} = 9 \left( \frac{1}{\sqrt{3}} \right)^2 - 1 = 3 - 1 = 2 \), which is also positive, indicating a minimum.

Step 6: Conclusion.
Thus, both points \( x = -\frac{1}{\sqrt{3}} \) and \( x = \frac{1}{\sqrt{3}} \) correspond to minimum values of the curve.