Question

The abscissae of the points of the curve $y = x^{3}$ in the interval [–2, 2], where the slope of the tangent can be obtained by mean value theorem for the interval [– 2, 2] are

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

Answer 1

Answer: (A)

$\pm \frac{2}{\sqrt{3}}$

Answer 2

Given that equation of curve $y = x^{3} = f(x)$

So $f(2) = 8$ and $f( - 2) = - 8$

Now $f^{'}(x) = 3x^{2}$ ⇒ $f^{'}(x) = \frac{f(2) - f( - 2)}{2 - ( - 2)}$

⇒$\frac{8 - ( - 8)}{4} = 3x^{2};\therefore x = \pm \frac{2}{\sqrt{3}}$.