Question

The maximum slope of the curve y = -x³ + 3x² + 2x -27 is

• A. 1
• B. 23
• C. 5
• D. -23

Answer 1

The Correct Option is C

We are asked to find the maximum slope of the curve defined by the equation:

\[ y = -x^3 + 3x^2 + 2x - 27 \]

Step 1: Find the slope function.

The slope of the curve at any point \( x \) is given by the first derivative of \( y \) with respect to \( x \), denoted as \( \frac{dy}{dx} \). Let's call the slope function \( S(x) \).

\[ S(x) = \frac{dy}{dx} = \frac{d}{dx} (-x^3 + 3x^2 + 2x - 27) \] \[ S(x) = -3x^2 + 3(2x) + 2 - 0 \] \[ S(x) = -3x^2 + 6x + 2 \]

So, the slope of the curve at any point \( x \) is given by the quadratic function \( S(x) = -3x^2 + 6x + 2 \).

Step 2: Find the maximum value of the slope function \( S(x) \).

The function \( S(x) = -3x^2 + 6x + 2 \) is a quadratic function representing a parabola. Since the coefficient of the \( x^2 \) term (\( a = -3 \)) is negative, the parabola opens downwards, and thus it has a maximum value at its vertex.

The x-coordinate of the vertex of a parabola \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \).

In our slope function \( S(x) \), \( a = -3 \) and \( b = 6 \). The value of \( x \) where the slope is maximum is:

\( x_{{max\ slope}}\) = \(-\frac{6}{2(-3)} \)= \(-\frac{6}{-6} \)= \(1\)

Alternatively, we can find the maximum by finding the critical points of \( S(x) \). We differentiate \( S(x) \) with respect to \( x \) and set the derivative \( S'(x) \) to zero.

\[ S'(x) = \frac{d}{dx}(-3x^2 + 6x + 2) = -6x + 6 \]

Set \( S'(x) = 0 \):

\[ -6x + 6 = 0 \] \[ -6x = -6 \] \[ x = 1 \]

To confirm this is a maximum, we check the second derivative \( S''(x) \):

\[ S''(x) = \frac{d}{dx}(-6x + 6) = -6 \]

Since \( S''(1) = -6 < 0 \), the slope \( S(x) \) has a maximum value at \( x = 1 \).

Step 3: Calculate the maximum slope value.

Substitute \( x = 1 \) into the slope function \( S(x) \):

\[ S_{\text{max}} = S(1) = -3(1)^2 + 6(1) + 2 \] \[ S_{\text{max}} = -3(1) + 6 + 2 \] \[ S_{\text{max}} = -3 + 6 + 2 \] \[ S_{\text{max}} = 5 \]

The maximum slope of the curve is 5.

So, the correct answer is (C): 5.

Answer 2

Given: 
Function: $y = -x^3 + 3x^2 + 2x - 27$

Step 1: Differentiate the function to find the slope
The slope of the curve at any point is given by the first derivative:
$$ \frac{dy}{dx} = y' = \frac{d}{dx}(-x^3 + 3x^2 + 2x - 27) = -3x^2 + 6x + 2 $$ 
Step 2: Find the maximum value of the slope
To find the maximum slope, we differentiate $y'$ with respect to $x$:
$$ \frac{d^2y}{dx^2} = y'' = \frac{d}{dx}(-3x^2 + 6x + 2) = -6x + 6 $$ Set $y'' = 0$ to find the critical point of the slope:
$$ -6x + 6 = 0 \Rightarrow x = 1 $$ 
Step 3: Verify whether this is a maximum
Check the second derivative around $x = 1$:
- For $x < 1$, $y'' > 0$ (function is concave up)
- For $x > 1$, $y'' < 0$ (function is concave down)
So, at $x = 1$, $y'$ attains a maximum value.

Step 4: Calculate the maximum slope
Substitute $x = 1$ into $y'$:
$$ y'(1) = -3(1)^2 + 6(1) + 2 = -3 + 6 + 2 = 5 $$ 
Final Answer:
Maximum slope = 5