Question

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

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

Hint:

Find c where f'(c) equals the average slope over the interval.

Answer 1

The Correct Option is C

Step 1: Chord slope
The chord joining $(-2,-8)$ and $(2,8)$ has slope 4.

Step 2: Tangent slope
The tangent slope is $3x^2$. Setting $3x^2=4$ gives $x^2=\dfrac{4}{3}$.

Step 3: Check
$\dfrac{2}{\sqrt3}\approx1.155$, which lies within $[-2,2]$, and so does its negative.

Final Answer:
The points have x equal to plus or minus 2/sqrt(3). \[ \boxed{\text{(C) }\pm\dfrac{2}{\sqrt{3}}} \]