Let f(1) = -2 and f'(x) ge 4.2 for 1 le x le | vedclass

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(D) According to the Lagrange's Mean Value Theorem $(LMVT)$,for a function $f(x)$ continuous on $[1, 6]$ and differentiable on $(1, 6)$,there exists a

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$19$

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(D) According to the Lagrange's Mean Value Theorem $(LMVT)$,for a function $f(x)$ continuous on $[1, 6]$ and differentiable on $(1, 6)$,there exists at least one $c \in (1, 6)$ such that:$f'(c) = \frac{f(6) - f(1)}{6 - 1}$Given $f(1) = -2$ and $f'(x) \ge 4.2$ for all $x \in [1, 6]$,we have:$f'(c) = \frac{f(6) - (-2)}{5} = \frac{f(6) + 2}{5}$Since $f'(c) \ge 4.2$,we can write:$\frac{f(6) + 2}{5} \ge 4.2$Multiplying both sides by $5$:$f(6) + 2 \ge 21$$f(6) \ge 19$Therefore,the smallest possible value of $f(6)$ is $19$.

Let $f(1) = -2$ and $f'(x) \ge 4.2$ for $1 \le x \le 6$. The smallest possible value of $f(6)$ is:

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