Let f(x) = begin{cases} g(x) cos(frac{1}{x}) | vedclass

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

Question

(B) By the definition of the derivative at $x = 0$:$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{g(h) \cos(\frac{1}{h}) - 0}{h}

Vedclass · Accepted Answer

is equal to $0$

Vedclass · Answer

(B) By the definition of the derivative at $x = 0$:$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{g(h) \cos(\frac{1}{h}) - 0}{h} = \lim_{h 	o 0} \frac{g(h)}{h} \cos(\frac{1}{h})$.Since $g(x)$ is an even function,$g(-x) = g(x)$.Given $g(x)$ is differentiable at $x = 0$ and passes through the origin,we have $g(0) = 0$.Thus,$g'(0) = \lim_{h 	o 0} \frac{g(h) - g(0)}{h} = \lim_{h 	o 0} \frac{g(h)}{h}$.Since $g(x)$ is even,$g'(0) = \lim_{h 	o 0} \frac{g(h) - g(0)}{h} = \lim_{h 	o 0} \frac{g(-h) - g(0)}{-h} = -g'(0)$,which implies $g'(0) = 0$.Therefore,$f'(0) = \lim_{h 	o 0} \frac{g(h)}{h} \cos(\frac{1}{h}) = g'(0) 	imes (	ext{a bounded value}) = 0 	imes 	ext{bounded value} = 0$.

Let $f(x) = \begin{cases} g(x) \cos(\frac{1}{x}) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$ where $g(x)$ is an even function differentiable at $x = 0$,passing through the origin. Then $f'(0)$:

Similar Questions

Let $f(x) = \begin{cases} \sin x, & \text{for } x \ge 0 \\ 1 - \cos x, & \text{for } x \le 0 \end{cases}$ and $g(x) = e^x$. Then $(g \circ f)'(0)$ is

If $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x = 0 \end{cases}$, then at $x = 2$, $f(x)$ is

If $f(x) = \begin{cases} ax^2 + b; & x \le 0 \\ x^2; & x > 0 \end{cases}$ possesses a derivative at $x = 0$,then:

If $f(x) = \begin{cases} \frac{x^2 \ln \cos x}{\ln(1 + x^2)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

The set of points where $f(x) = \frac{x}{4+|x|}$ is differentiable is

Vedclass Test Series

Exam Paper Generator

Online Exam Module