Question

Find the maximum slope of the curve \( y = x^3 + 3x^2 + 9x - 30 \).

Hint:

To find the maximum slope, differentiate the function to get \( \frac{dy}{dx} \), then find critical points using the second derivative. If the slope increases indefinitely, check the problem context for intended bounds.

Answer 1

The slope of the curve \( y = x^3 + 3x^2 + 9x - 30 \) at any point is given by the first derivative \( \frac{dy}{dx} \), which represents the rate of change of \( y \) with respect to \( x \). To find the maximum slope, we need to maximize \( \frac{dy}{dx} \). Step 1: Compute the First Derivative
Differentiate \( y = x^3 + 3x^2 + 9x - 30 \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} (x^3) + \frac{d}{dx} (3x^2) + \frac{d}{dx} (9x) - \frac{d}{dx} (30) \] \[ \frac{dy}{dx} = 3x^2 + 6x + 9 \] Step 2: Find the Maximum Value of the Slope
To maximize \( \frac{dy}{dx} = 3x^2 + 6x + 9 \), we take the second derivative and find the critical points where the slope of \( \frac{dy}{dx} \) is zero (i.e., where \( \frac{dy}{dx} \) has a maximum or minimum). Differentiate \( \frac{dy}{dx} = 3x^2 + 6x + 9 \) with respect to \( x \): \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} (3x^2 + 6x + 9) = 6x + 6 \] Set the second derivative equal to zero to find critical points: \[ 6x + 6 = 0 \] \[ 6x = -6 \] \[ x = -1 \] Step 3: Determine the Nature of the Critical Point
Use the second derivative test to check if \( x = -1 \) is a maximum: - If \( \frac{d^2 y}{dx^2} < 0 \), the function has a local maximum. - If \( \frac{d^2 y}{dx^2} > 0 \), the function has a local minimum. Substitute \( x = -1 \) into \( \frac{d^2 y}{dx^2} \): \[ \frac{d^2 y}{dx^2} \bigg|_{x = -1} = 6(-1) + 6 = -6 + 6 = 0 \] Since \( \frac{d^2 y}{dx^2} = 0 \) at \( x = -1 \), the second derivative test is inconclusive. We need to check the first derivative around \( x = -1 \) or use the first derivative test. Step 4: First Derivative Test
Evaluate \( \frac{dy}{dx} = 3x^2 + 6x + 9 \) around \( x = -1 \): - For \( x = -2 \): \[ \frac{dy}{dx} = 3(-2)^2 + 6(-2) + 9 = 12 - 12 + 9 = 9 \] - For \( x = 0 \): \[ \frac{dy}{dx} = 3(0)^2 + 6(0) + 9 = 9 \] - For \( x = -1 \): \[ \frac{dy}{dx} = 3(-1)^2 + 6(-1) + 9 = 3 - 6 + 9 = 6 \] The calculation seems inconsistent. Let's compute \( \frac{dy}{dx} \) correctly at \( x = -1 \): \[ \frac{dy}{dx} = 3(-1)^2 + 6(-1) + 9 = 3 - 6 + 9 = 6 \] Let's try points closer to check the behavior: - For \( x = -1.5 \): \[ \frac{dy}{dx} = 3(-1.5)^2 + 6(-1.5) + 9 = 3(2.25) - 9 + 9 = 6.75 \] - For \( x = -0.5 \): \[ \frac{dy}{dx} = 3(-0.5)^2 + 6(-0.5) + 9 = 3(0.25) - 3 + 9 = 0.75 + 6 = 6.75 \] This indicates a minimum, not a maximum. Let's find the critical point by setting the first derivative to zero:
\[ 3x^2 + 6x + 9 = 0 \] Divide by 3: \[ x^2 + 2x + 3 = 0 \] Discriminant: \[ \Delta = 2^2 - 4(1)(3) = 4 - 12 = -8 \] Since the discriminant is negative, \( x^2 + 2x + 3 = 0 \) has no real roots, meaning \( \frac{dy}{dx} = 3x^2 + 6x + 9>0 \) for all \( x \) (as the quadratic opens upwards and the minimum value is positive). Step 5: Re-evaluate the Maximum Slope
Since \( \frac{dy}{dx} = 3x^2 + 6x + 9 \) is a quadratic in \( x \) with a positive leading coefficient, it has a minimum value (not a maximum) at its vertex. The vertex of \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \): \[ x = -\frac{6}{2 \cdot 3} = -1 \] At \( x = -1 \): \[ \frac{dy}{dx} = 6 \] As \( x \to \pm \infty \), \( 3x^2 + 6x + 9 \to \infty \).
Let's check the behavior:
- For \( x = 10 \): \[ \frac{dy}{dx} = 3(10)^2 + 6(10) + 9 = 300 + 60 + 9 = 369 \] - For \( x = -10 \): \[ \frac{dy}{dx} = 3(-10)^2 + 6(-10) + 9 = 300 - 60 + 9 = 249 \] The slope increases without bound as \( |x| \) increases. However, the problem asks for the maximum slope, which suggests a possible misinterpretation. Let's assume the intent is to find the maximum value of the slope function within a practical context or correct the function. The slope \( \frac{dy}{dx} \) has no upper bound, but let's derive correctly.