Let a 0 and f be continuous in [-a, a]. Sup | vedclass

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(A) By the Lagrange's Mean Value Theorem $(LMVT)$ on the interval $[-a, 0]$,there exists some $c_1 \in (-a, 0)$ such that $f'(c_1) = \frac{f(0) - f(-a

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equals $0$

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(A) By the Lagrange's Mean Value Theorem $(LMVT)$ on the interval $[-a, 0]$,there exists some $c_1 \in (-a, 0)$ such that $f'(c_1) = \frac{f(0) - f(-a)}{0 - (-a)} = \frac{f(0) + a}{a}$.Since $f'(c_1) \le 1$,we have $\frac{f(0) + a}{a} \le 1$,which implies $f(0) + a \le a$,so $f(0) \le 0$.By the $LMVT$ on the interval $[0, a]$,there exists some $c_2 \in (0, a)$ such that $f'(c_2) = \frac{f(a) - f(0)}{a - 0} = \frac{a - f(0)}{a}$.Since $f'(c_2) \le 1$,we have $\frac{a - f(0)}{a} \le 1$,which implies $a - f(0) \le a$,so $-f(0) \le 0$,or $f(0) \ge 0$.Since $f(0) \le 0$ and $f(0) \ge 0$,it must be that $f(0) = 0$.

Let $a > 0$ and $f$ be continuous in $[-a, a]$. Suppose that $f'(x)$ exists and $f'(x) \le 1$ for all $x \in (-a, a)$. If $f(a) = a$ and $f(-a) = -a$,then $f(0)$ is:

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