Suppose that f(x) is a differentiable functio | vedclass

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

Question

(C) We are given $g(x) = x f^{\prime}(x)$.To find $g^{\prime}(0)$, we use the definition of the derivative:$g^{\prime}(0) = \lim_{h 	o 0} \frac{g(0+h

Vedclass · Accepted Answer

$g^{\prime}(0)=1$

Vedclass · Answer

(C) We are given $g(x) = x f^{\prime}(x)$.To find $g^{\prime}(0)$, we use the definition of the derivative:$g^{\prime}(0) = \lim_{h 	o 0} \frac{g(0+h) - g(0)}{h}$.Since $g(0) = 0 \cdot f^{\prime}(0) = 0$, we have:$g^{\prime}(0) = \lim_{h 	o 0} \frac{h f^{\prime}(h) - 0}{h} = \lim_{h 	o 0} f^{\prime}(h)$.Since $f^{\prime}(x)$ is given to be continuous at $x=0$, we have $\lim_{h 	o 0} f^{\prime}(h) = f^{\prime}(0)$.Given $f^{\prime}(0) = 1$, it follows that $g^{\prime}(0) = 1$.

Suppose that $f(x)$ is a differentiable function such that $f^{\prime}(x)$ is continuous, $f^{\prime}(0)=1$ and $f^{\prime \prime}(0)$ does not exist. Let $g(x)=x f^{\prime}(x)$. Then,

Similar Questions

If $f(x)=|x|+|\sin x|$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,then its left hand derivative at $x=0$ is

The number of points in the interval $(0,2)$ at which $f(x)=|x-0.5|+|x-1|+\tan x$ is not differentiable is

If $\alpha \in R - \{-1\}$ and $f(x) = |(|x| + \alpha)(|x| - 1)|$,then the number of points at which $f(x)$ is not differentiable is:

If $f(x) = |x - 3|,$ then $f$ is

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module