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.
Define \( f(x) = \begin{cases} x^2 + bx + c, & x< 1 \\ x, & x \geq 1 \end{cases} \). If f(x) is differentiable at x=1, then b−c is equal to