Let f(x) be continuous on [0, 5] and differen | vedclass

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(A) By the Mean Value Theorem, for any $x \in (0, 5]$, there exists a $c \in (0, x)$ such that $f^{\prime}(c) = \frac{f(x) - f(0)}{x - 0}$.Given $f(0)

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$|f(x)| \leq 1$

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(A) By the Mean Value Theorem, for any $x \in (0, 5]$, there exists a $c \in (0, x)$ such that $f^{\prime}(c) = \frac{f(x) - f(0)}{x - 0}$.Given $f(0) = 0$, we have $f^{\prime}(c) = \frac{f(x)}{x}$.Taking the absolute value, $|f^{\prime}(c)| = \left|\frac{f(x)}{x}\right| = \frac{|f(x)|}{|x|}$.Since $|f^{\prime}(x)| \leq \frac{1}{5}$ for all $x \in (0, 5)$, it follows that $|f^{\prime}(c)| \leq \frac{1}{5}$.Therefore, $\frac{|f(x)|}{x} \leq \frac{1}{5}$, which implies $|f(x)| \leq \frac{x}{5}$.Since $x \in [0, 5]$, the maximum value of $\frac{x}{5}$ is $\frac{5}{5} = 1$.Thus, $|f(x)| \leq 1$ for all $x \in [0, 5]$.

Let $f(x)$ be continuous on $[0, 5]$ and differentiable in $(0, 5)$. If $f(0) = 0$ and $|f^{\prime}(x)| \leq \frac{1}{5}$ for all $x$ in $(0, 5)$, then which of the following is true for all $x$ in $[0, 5]$?

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