Question

The slope of the curve \( y = -x^3 + 3x^2 + 8x - 20 \) is maximum at:

• A. \( (1, -10) \)
• B. \( (1, 10) \)
• C. \( (10, 1) \)
• D. \( (-10, 1) \)

Hint:

To find the maximum slope, always take the first derivative and solve for critical points. Then, check the second derivative to confirm if the point is a maximum or minimum.

Answer 1

The Correct Option is B

To find the point where the slope is maximum, we need to take the derivative of the function \( y = -x^3 + 3x^2 + 8x - 20 \) with respect to \( x \) and find the critical points. First, compute \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -3x^2 + 6x + 8. \] Next, set the first derivative equal to zero to find the critical points: \[ -3x^2 + 6x + 8 = 0. \] Solving this quadratic equation gives: \[ x = 1 \quad \text{(as the maximum value is at \( x = 1 \))}. \] Substituting \( x = 1 \) into the original equation to find the corresponding \( y \)-value: \[ y = -(1)^3 + 3(1)^2 + 8(1) - 20 = -1 + 3 + 8 - 20 = -10. \] Hence, the slope is maximum at \( (1, -10) \).