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(B) Given $|f^{\prime}(x)| \leq 5$ for all $x \in \mathbb{R}$.
By the Mean Value Theorem, for any $x$, there exists $c$ between $0$ and $x$ such that

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$[-5, 5]$

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(B) Given $|f^{\prime}(x)| \leq 5$ for all $x \in \mathbb{R}$.
By the Mean Value Theorem, for any $x$, there exists $c$ between $0$ and $x$ such that $f(x) - f(0) = f^{\prime}(c)(x - 0)$.
Since $f(0) = 0$, we have $f(x) = f^{\prime}(c) \cdot x$.
For $x = 1$, $f(1) = f^{\prime}(c) \cdot 1 = f^{\prime}(c)$.
Since $|f^{\prime}(c)| \leq 5$, it follows that $|f(1)| \leq 5$.
Alternatively, using integration:
$\int_{0}^{1} -5 \, dx \leq \int_{0}^{1} f^{\prime}(x) \, dx \leq \int_{0}^{1} 5 \, dx$
$-5 \leq f(1) - f(0) \leq 5$
Since $f(0) = 0$, we get $-5 \leq f(1) \leq 5$.
Thus, $f(1) \in [-5, 5]$.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be such that $f(0)=0$ and $|f^{\prime}(x)| \leq 5$ for all $x$. Then $f(1)$ is in

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