If f(x) is a differentiable function such tha | vedclass

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

Question

(B) Given that $f\left( \frac{1}{n} \right) = 0$ for all $n \in \{1, 2, 3, \dots\}$.Since $f(x)$ is continuous (as it is differentiable),we have $f(0)

Vedclass · Accepted Answer

$f(0) = 0$ and $f'(0) = 0$

Vedclass · Answer

(B) Given that $f\left( \frac{1}{n} ight) = 0$ for all $n \in \{1, 2, 3, \dots\}$.Since $f(x)$ is continuous (as it is differentiable),we have $f(0) = \lim_{n 	o \infty} f\left( \frac{1}{n} ight) = 0$.Now,consider the sequence $x_n = \frac{1}{n}$. We have $f(x_n) = 0$ and $f(0) = 0$.By the definition of the derivative at $x = 0$:$f'(0) = \lim_{x 	o 0} \frac{f(x) - f(0)}{x - 0} = \lim_{n 	o \infty} \frac{f(1/n) - f(0)}{1/n - 0} = \lim_{n 	o \infty} \frac{0 - 0}{1/n} = 0$.Thus,$f(0) = 0$ and $f'(0) = 0$.

If $f(x)$ is a differentiable function such that $f: R \to R$ and $f\left( \frac{1}{n} \right) = 0$ for all $n \ge 1, n \in I$,then:

Similar Questions

If $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$,then $f^{\prime}(1)$ is equal to:

The set of all points,where the derivative of the function $f(x) = \frac{x}{1+|x|}$ exists,is

If $[t]$ denotes the greatest integer $\leq t$,then the number of points at which the function $f(x) = 4|2x + 3| + 9[x + \frac{1}{2}] - 12[x + 20]$ is not differentiable in the open interval $(-20, 20)$ is:

Given $f(x) = \begin{cases} 1 + x & x < 0 \\ 2 - 3x & x \geq 0 \end{cases}$,find the critical point $x = \dots \dots$.

Let $K$ be the set of all real values of $x$ where the function $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$ is not differentiable. Then the set $K$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module