Question

The minimum value of the slope of the tangent to curve $y = x^3 - 3x^2 + 2x + 93$ is

• A. 1
• B. -1
• C. 2
• D. -2

Hint:

For a quadratic $ax^2 + bx + c$, the minimum value is reached at $x = -b/2a$ and the value is $-\Delta/4a$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The slope of a tangent is the first derivative of the function. We need to find the global minimum of this derivative.
Step 3: Detailed Explanation:
1. Let \( f(x) = y = x^3 - 3x^2 + 2x + 93 \).
2. Slope \( m = \frac{dy}{dx} = 3x^2 - 6x + 2 \).
3. To find the minimum of \( m \), differentiate \( m \) with respect to \( x \):
\[ \frac{dm}{dx} = 6x - 6 \]
4. Set \( \frac{dm}{dx} = 0 \implies 6x = 6 \implies x = 1 \).
5. Second derivative check: \( \frac{d^2m}{dx^2} = 6 \). Since \( 6>0 \), \( x = 1 \) is a point of local minimum for the slope.
6. Minimum slope value at \( x = 1 \):
\[ m_{min} = 3(1)^2 - 6(1) + 2 = 3 - 6 + 2 = -1 \]
Step 4: Final Answer:
The minimum value of the slope is \( -1 \).