Let f: R rightarrow R be given byf(x) = begin | vedclass

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(B) First,we analyze the function $f(x)$ in different intervals.For $x < 0$,$f(x) = (x+1)^5 - 2x$. As $x \rightarrow -\infty$,$f(x) \rightarrow -\inft

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$1, 2, 4$

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(B) First,we analyze the function $f(x)$ in different intervals.For $x $f^{\prime}(x) = 5(x+1)^4 - 2$. Setting $f^{\prime}(x) = 0$ gives $(x+1)^4 = 2/5$,so $x = -1 \pm (2/5)^{1/4}$. Since $f^{\prime}(x)$ changes sign in $(-\infty, 0)$,$f(x)$ is not monotonic on $(-\infty, 0)$. Thus,statement $(3)$ is incorrect.For $x \geq 3$,$f(x)$ is continuous,$f(3) = 1/3$,and $\lim_{x \rightarrow \infty} f(x) = \infty$. The range is $[1/3, \infty)$.Combining the ranges of all intervals,we find that the range of $f(x)$ is $R$,so $f$ is onto. Statement $(2)$ is correct.Now,consider $f^{\prime}(x)$ for $x$ near $1$:For $0 \leq x For $1 \leq x For $x$ slightly greater than $1$,$f^{\prime}(x) = 2x^2 - 8x + 7$. The derivative of $f^{\prime}(x)$ is $f^{\prime\prime}(x) = 4x - 8$. At $x=1$,$f^{\prime\prime}(1^+) = -4$.Since the left-hand derivative of $f^{\prime}$ at $x=1$ is $2$ and the right-hand derivative is $-4$,$f^{\prime}$ is not differentiable at $x=1$. Statement $(4)$ is correct.Also,since $f^{\prime}(x)$ increases to $1$ at $x=1$ and decreases for $x > 1$,$f^{\prime}$ has a local maximum at $x=1$. Statement $(1)$ is correct.Therefore,statements $(1), (2),$ and $(4)$ are correct.

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