For the function f(x) = x cos frac{1}{x}, qua | vedclass

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(A) Given $f(x) = x \cos \frac{1}{x}$ for $x \geq 1$.First,find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = \cos \frac{1}{x} + x \left( -\sin \fra

Vedclass · Accepted Answer

$(B, C, D)$

Vedclass · Answer

(A) Given $f(x) = x \cos \frac{1}{x}$ for $x \geq 1$.First,find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = \cos \frac{1}{x} + x \left( -\sin \frac{1}{x} ight) \left( -\frac{1}{x^2} ight) = \cos \frac{1}{x} + \frac{1}{x} \sin \frac{1}{x}$.Now,evaluate $\lim _{x ightarrow \infty} f^{\prime}(x)$:$\lim _{x ightarrow \infty} \left( \cos \frac{1}{x} + \frac{1}{x} \sin \frac{1}{x} ight) = \cos(0) + 0 \cdot \sin(0) = 1 + 0 = 1$. Thus,statement $(B)$ is correct.Next,find the second derivative $f^{\prime \prime}(x)$:$f^{\prime \prime}(x) = -\sin \frac{1}{x} \left( -\frac{1}{x^2} ight) + \left( -\frac{1}{x^2} ight) \sin \frac{1}{x} + \frac{1}{x} \cos \frac{1}{x} \left( -\frac{1}{x^2} ight) = \frac{1}{x^2} \sin \frac{1}{x} - \frac{1}{x^2} \sin \frac{1}{x} - \frac{1}{x^3} \cos \frac{1}{x} = -\frac{1}{x^3} \cos \frac{1}{x}$.For $x \in [1, \infty)$,$\frac{1}{x} \in (0, 1]$. Since $\cos 	heta > 0$ for $	heta \in (0, 1]$,$f^{\prime \prime}(x) = -\frac{1}{x^3} \cos \frac{1}{x} By the Mean Value Theorem $(LMVT)$ on $[x, x+2]$,there exists $c \in (x, x+2)$ such that $\frac{f(x+2)-f(x)}{2} = f^{\prime}(c)$.Since $f^{\prime}(x)$ is strictly decreasing and $\lim _{x ightarrow \infty} f^{\prime}(x) = 1$,we have $f^{\prime}(x) > 1$ for all $x \in [1, \infty)$.Therefore,$f^{\prime}(c) > 1$,which implies $\frac{f(x+2)-f(x)}{2} > 1$,or $f(x+2)-f(x) > 2$. Thus,statement $(C)$ is correct and $(A)$ is incorrect.The correct combination is $(B, C, D)$.

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