Question

Maximum slope of the curve \(y = -2x^3 + 6x^2 + 5x - 20\) is:

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The Correct Option is C

To find the maximum slope of the curve \(y = -2x^3 + 6x^2 + 5x - 20\), we must determine when the derivative reaches its maximum value. The derivative \(\frac{dy}{dx}\) gives us the slope of the tangent at any point on the curve.

First, we find the derivative: \(\frac{dy}{dx} = \frac{d}{dx}(-2x^3 + 6x^2 + 5x - 20)\). Using basic differentiation rules, we get:

\(\frac{dy}{dx} = -6x^2 + 12x + 5\).

Next, locate the critical points of the derivative by setting the second derivative to zero, as these points indicate maximum or minimum slopes:

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(-6x^2 + 12x + 5)\).

This results in: \(-12x + 12\).

Set the second derivative to zero to find critical points: \(-12x + 12 = 0\). Solving gives: \(x = 1\).

To determine if this point is a maximum, calculate the second derivative: \(\frac{d^2y}{dx^2} = -12 < 0\), which indicates a maximum.

Substitute \(x = 1\) into the first derivative to find the maximum slope:

\(\frac{dy}{dx}\bigg|_{x=1} = -6(1)^2 + 12(1) + 5 = -6 + 12 + 5 = 11\).

Therefore, the maximum slope of the curve is 11.