On the curve y=x^3,the point at which the tan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $f(x) = x^3$. The slope of the chord joining the points $(-1, -1)$ and $(2, 8)$ is given by $m = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{8 - (-1

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(C) Let $f(x) = x^3$. The slope of the chord joining the points $(-1, -1)$ and $(2, 8)$ is given by $m = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{8 - (-1)}{3} = \frac{9}{3} = 3$. Since the tangent line is parallel to this chord,the derivative $f'(x)$ at the point must be equal to the slope of the chord. $f'(x) = \frac{d}{dx}(x^3) = 3x^2$. Setting $f'(x) = 3$,we get $3x^2 = 3$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$. For $x = 1$,$y = (1)^3 = 1$,giving the point $(1, 1)$. For $x = -1$,$y = (-1)^3 = -1$,giving the point $(-1, -1)$. Since $(-1, -1)$ is one of the endpoints of the chord,the point on the curve between the given points is $(1, 1)$.

On the curve $y=x^3$,the point at which the tangent line is parallel to the chord joining the points $(-1, -1)$ and $(2, 8)$ is

Similar Questions

The equation of the normal drawn to the curve $y^3=4 x^5$ at the point $(4,16)$ is

Find the points on the curve $y=x^{3}$ at which the slope of the tangent is equal to the $y$-coordinate of the point.

The sum of the lengths of the subtangent and the subnormal drawn at $\theta = \frac{\pi}{3}$ on the cycloid $x = a(\theta - \sin \theta)$,$y = a(1 - \cos \theta)$ is

The normal at the point $(1,1)$ on the curve $2y + x^{2} = 3$ is

If the tangent to the curve $xy + ax + by = 0$ at $(1, 1)$ makes an angle $\tan^{-1}(2)$ with the $x$-axis,then $\left( \frac{a + b}{ab} \right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module