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

If Rolle's theorem holds for the function $f(x) = 2x^3 + bx^2 + cx$ on the interval $x \in [-1, 1]$ at the point $x = \frac{1}{2}$,then the value of $2b + c$ is:

Consider $f(x) = |1 - x|$ for $1 \le x \le 2$ and $g(x) = f(x) + b \sin(\frac{\pi}{2}x)$ for $1 \le x \le 2$. Which of the following is correct?

The value of $C$ in the Mean Value Theorem for the function $f(x) = x^{2}$ in the interval $[2, 4]$ is:

Consider the function $f(x) = |x - 2| + |x - 5|, x \in R$.
Statement-$1$: $f'(4) = 0$.
Statement-$2$: $f$ is continuous in $[2, 5]$,differentiable in $(2, 5)$ and $f(2) = f(5)$.

If $f(x) = ax^3 + bx^2 + 11x - 6$ for $x \in [1, 3]$ satisfies the conditions of Rolle's theorem and $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$,find $a$ and $b$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module