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 Question:
The Mean Value Theorem (MVT) states that for a function \( f(x) \) continuous on \( [a, b] \) and differentiable on \( (a, b) \), there exists at least one point \( c \in (a, b) \) such that the slope of the tangent at \( c \) is equal to the average slope of the function over the interval.
Step 2: Key Formula or Approach:
The formula for Mean Value Theorem is:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \] Given function \( f(x) = x^3 \) and interval \( [-2, 2] \), so \( a = -2 \) and \( b = 2 \).
Step 3: Detailed Explanation:
First, calculate the derivative of the function:
\[ f'(x) = 3x^2 \] Next, calculate the value of the function at the endpoints:
\[ f(2) = (2)^3 = 8 \]
\[ f(-2) = (-2)^3 = -8 \]
Apply the MVT formula:
\[ f'(c) = \frac{8 - (-8)}{2 - (-2)} = \frac{16}{4} = 4 \]
Set the derivative equal to this value:
\[ 3c^2 = 4 \]
\[ c^2 = \frac{4}{3} \]
\[ c = \pm \frac{2}{\sqrt{3}} \]
Both values \( \frac{2}{\sqrt{3}} \) and \( -\frac{2}{\sqrt{3}} \) lie within the interval \( (-2, 2) \).
Step 4: Final Answer:
The abscissae are \( \pm \frac{2}{\sqrt{3}} \).