If f(x) = begin{cases} x^3 sin left(frac{1}{x | vedclass

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

Question

(B) First,we find $f^{\prime}(0)$ using the definition of the derivative: $f^{\prime}(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac

Vedclass · Accepted Answer

$f^{\prime \prime}\left(\frac{2}{\pi}\right) = \frac{24-\pi^2}{2 \pi}$

Vedclass · Answer

(B) First,we find $f^{\prime}(0)$ using the definition of the derivative: $f^{\prime}(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^3 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h^2 \sin(1/h) = 0$. Next,for $x 
eq 0$,$f^{\prime}(x) = 3x^2 \sin(1/x) - x \cos(1/x)$. Now,we find $f^{\prime \prime}(0)$ using the definition: $f^{\prime \prime}(0) = \lim_{h 	o 0} \frac{f^{\prime}(h) - f^{\prime}(0)}{h} = \lim_{h 	o 0} \frac{3h^2 \sin(1/h) - h \cos(1/h) - 0}{h} = \lim_{h 	o 0} (3h \sin(1/h) - \cos(1/h))$. Since $\lim_{h 	o 0} 3h \sin(1/h) = 0$ but $\lim_{h 	o 0} \cos(1/h)$ does not exist,$f^{\prime \prime}(0)$ does not exist. For $x 
eq 0$,$f^{\prime \prime}(x) = \frac{d}{dx} [3x^2 \sin(1/x) - x \cos(1/x)] = 6x \sin(1/x) - 3 \cos(1/x) - (\cos(1/x) + x(-\sin(1/x))(-1/x^2)) = 6x \sin(1/x) - 4 \cos(1/x) - \frac{1}{x} \sin(1/x)$. Evaluating at $x = \frac{2}{\pi}$: $f^{\prime \prime}\left(\frac{2}{\pi}ight) = 6(\frac{2}{\pi}) \sin(\frac{\pi}{2}) - 4 \cos(\frac{\pi}{2}) - \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{12}{\pi}(1) - 4(0) - \frac{\pi}{2}(1) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24-\pi^2}{2 \pi}$.

If $f(x) = \begin{cases} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}$,then

Similar Questions

If the three functions $f(x)$,$g(x)$,and $h(x)$ are such that $h(x) = f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x) = c$,where $c$ is a constant,then $\frac{f^{\prime \prime}(x)}{f(x)} + \frac{g^{\prime \prime}(x)}{g(x)} + \frac{2c}{f(x) \cdot g(x)}$ is equal to:

Let $f$ be a twice differentiable function such that $f^{\prime \prime}(x) = -f(x)$,$f^{\prime}(x) = g(x)$,and $h(x) = (f(x))^2 + (g(x))^2$. If $h(5) = 1$,then the value of $h(10)$ is

If $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$ and $f(0) = 1$,then $f(x)$ is equal to

If $y=5 \cos x-3 \sin x$,then $\frac{d^2 y}{d x^2}=$ . . . . . . .

If $\sqrt{x+y}-\sqrt{x-y}=c$, then $\frac{d^2 y}{d x^2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module