The abscissa of the points of the curve y = x | vedclass

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

Question

(A) Given the curve $f(x) = x^3$ on the interval $[-2, 2]$.According to the Mean Value Theorem,there exists at least one point $c \in (-2, 2)$ such th

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the curve $f(x) = x^3$ on the interval $[-2, 2]$.According to the Mean Value Theorem,there exists at least one point $c \in (-2, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Here,$a = -2$ and $b = 2$.$f(2) = 2^3 = 8$ and $f(-2) = (-2)^3 = -8$.The slope of the secant line is $\frac{f(2) - f(-2)}{2 - (-2)} = \frac{8 - (-8)}{4} = \frac{16}{4} = 4$.The derivative is $f'(x) = 3x^2$.Setting $f'(c) = 4$,we get $3c^2 = 4$.$c^2 = \frac{4}{3}$,which implies $c = \pm \frac{2}{\sqrt{3}}$.

The abscissa of the points of the curve $y = x^3$ in the interval $[-2, 2]$,where the slope of the tangents is equal to the slope of the secant line as per the Mean Value Theorem for the interval $[-2, 2]$,are:

Similar Questions

Let $f: D \rightarrow R$ where $D=[0,1] \cup [2,4]$ be defined by $f(x)=\begin{cases} x, & \text{if } x \in [0,1] \\ 4-x, & \text{if } x \in [2,4] \end{cases}$. Then,

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. Then the values of $a$ and $b$ are respectively

Let $f:[1,3] \rightarrow R$ be a continuous function that is differentiable in $(1,3)$ and $f^{\prime}(x)=|f(x)|^{2}+4$ for all $x \in(1,3).$ Then,

Let $f:[1,3] \rightarrow R$ be continuous and differentiable in $(1,3)$ such that $f^{\prime}(x)=[f(x)]^2+4$ for all $x \in (1,3)$. Then:

In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$,and $f(x) = \sqrt{x}$,then the value of $c$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module