Let the function f:[-7,0] rightarrow R be con | vedclass

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(B) By the Mean Value Theorem $(LMVT)$ on the interval $[-7, -1]$,there exists some $c_1 \in (-7, -1)$ such that $\frac{f(-1) - f(-7)}{-1 - (-7)} = f'

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$(-\infty, 20]$

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(B) By the Mean Value Theorem $(LMVT)$ on the interval $[-7, -1]$,there exists some $c_1 \in (-7, -1)$ such that $\frac{f(-1) - f(-7)}{-1 - (-7)} = f'(c_1)$.Given $f'(x) \leq 2$,we have $\frac{f(-1) - (-3)}{6} \leq 2$,which implies $f(-1) + 3 \leq 12$,so $f(-1) \leq 9$.By the Mean Value Theorem $(LMVT)$ on the interval $[-7, 0]$,there exists some $c_2 \in (-7, 0)$ such that $\frac{f(0) - f(-7)}{0 - (-7)} = f'(c_2)$.Given $f'(x) \leq 2$,we have $\frac{f(0) - (-3)}{7} \leq 2$,which implies $f(0) + 3 \leq 14$,so $f(0) \leq 11$.Adding these two inequalities,we get $f(-1) + f(0) \leq 9 + 11 = 20$.Since $f(-1)$ and $f(0)$ can be arbitrarily small,the interval is $(-\infty, 20]$.

Let the function $f:[-7,0] \rightarrow R$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and $f'(x) \leq 2$ for all $x \in (-7,0)$,then for all such functions $f$,$f(-1)+f(0)$ lies in the interval:

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