Question

The abscissae of the points of the curve \[ y = x^3 \quad \text{are in the interval} \quad [-2, 2], \] where the slope of the tangents can be obtained by the mean value theorem for the interval \( [-2, 2] \), are

• A. 0
• B. \( \pm \sqrt{3} \)
• C. \( \frac{2}{3} \)
• D. \( \sqrt{2} \)

Hint:

The mean value theorem helps find the slope of the tangent at some point between two points of a function by calculating the average rate of change over the interval.

Answer 1

The Correct Option is B

Step 1: Understanding the mean value theorem.
The mean value theorem states that for a continuous function \( f(x) \) on the interval \( [a, b] \), there exists a point \( c \) in \( (a, b) \) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]
In this problem, the curve is given by \( y = x^3 \), and we are asked to find the slope of the tangent using the mean value theorem over the interval \( [-2, 2] \).

Step 2: Apply the mean value theorem.
First, we compute \( f(x) = x^3 \) at the endpoints of the interval:
\[ f(-2) = (-2)^3 = -8, \quad f(2) = (2)^3 = 8. \]
Using the mean value theorem:
\[ f'(c) = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{8 - (-8)}{4} = \frac{16}{4} = 4. \]

Step 3: Conclusion.
Thus, the slope of the tangent at the point \( c \) is 4, and the correct answer is \( \pm \sqrt{3} \).
Final Answer:
The correct answer is: \[ \boxed{\pm \sqrt{3}}. \]