Let f: R rightarrow R be defined as f(x) = be | vedclass

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(A) To find $f''(0)$,we first find $f'(x)$.For $x < 0$,$f'(x) = 5x^4 \sin(1/x) - x^3 \cos(1/x) + 10x$. Thus,$f'(0) = \lim_{h 	o 0^-} \frac{f(h) - f(0

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$5$

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(A) To find $f''(0)$,we first find $f'(x)$.For $x For $x > 0$,$f'(x) = 5x^4 \cos(1/x) + x^3 \sin(1/h) + 2\lambda x$. Thus,$f'(0) = \lim_{h 	o 0^+} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0^+} (h^4 \cos(1/h) + \lambda h) = 0$.Now,$f''(0) = \lim_{h 	o 0} \frac{f'(h) - f'(0)}{h}$.Left-hand derivative $(LHD)$ at $x=0$: $\lim_{h 	o 0^-} \frac{5h^4 \sin(1/h) - h^3 \cos(1/h) + 10h - 0}{h} = \lim_{h 	o 0^-} (5h^3 \sin(1/h) - h^2 \cos(1/h) + 10) = 10$.Right-hand derivative $(RHD)$ at $x=0$: $\lim_{h 	o 0^+} \frac{5h^4 \cos(1/h) + h^3 \sin(1/h) + 2\lambda h - 0}{h} = \lim_{h 	o 0^+} (5h^3 \cos(1/h) + h^2 \sin(1/h) + 2\lambda) = 2\lambda$.For $f''(0)$ to exist,$LHD$ = $RHD$,so $2\lambda = 10$,which implies $\lambda = 5$.

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} x^{5} \sin \left(\frac{1}{x}\right) + 5x^{2} & , x < 0 \\ 0 & , x = 0 \\ x^{5} \cos \left(\frac{1}{x}\right) + \lambda x^{2} & , x > 0 \end{cases}$. The value of $\lambda$ for which $f''(0)$ exists is:

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