Abscissa of points of curve y = {x^3} in interval [-2, 2] , where slope of tangents can be obtained

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5. Continuity and Differentiation

easy

The abscissa of the points of the curve $y = {x^3}$ 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

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

B

$ \pm \sqrt 3 $

C

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

D

$0$

Solution

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

So $f(2) = 8$ and $f( – 2) = – 8$

Now $f'(x) = 3{x^2} \Rightarrow f'(x) = \frac{{f(2) – f( – 2)}}{{2 – ( – 2)}}$

==>$\frac{{8 – ( – 8)}}{4} = 3{x^2};\,\,\,\,$

$\therefore x = \pm \frac{2}{{\sqrt 3 }}$.

Standard 12

Mathematics

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