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(C) Given $f(2)=8$,$f'(2)=5$,$f'(x) \geq 1$,and $f''(x) \geq 4$ for all $x \in (1,6)$.By the Mean Value Theorem on the interval $[2, 5]$ for the funct

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$f(5)+f'(5) \geq 28$

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(C) Given $f(2)=8$,$f'(2)=5$,$f'(x) \geq 1$,and $f''(x) \geq 4$ for all $x \in (1,6)$.By the Mean Value Theorem on the interval $[2, 5]$ for the function $f'(x)$,there exists $c \in (2, 5)$ such that $f''(c) = \frac{f'(5)-f'(2)}{5-2}$.Since $f''(x) \geq 4$,we have $\frac{f'(5)-5}{3} \geq 4 \Rightarrow f'(5)-5 \geq 12 \Rightarrow f'(5) \geq 17$.By the Mean Value Theorem on the interval $[2, 5]$ for the function $f(x)$,there exists $d \in (2, 5)$ such that $f'(d) = \frac{f(5)-f(2)}{5-2}$.Since $f'(x) \geq 1$,we have $\frac{f(5)-8}{3} \geq 1 \Rightarrow f(5)-8 \geq 3 \Rightarrow f(5) \geq 11$.However,we can use Taylor's theorem or integration: $f(5) = f(2) + \int_{2}^{5} f'(x) dx$.Since $f'(x) \geq f'(2) + \int_{2}^{x} f''(t) dt \geq 5 + 4(x-2) = 4x-3$.Then $f(5) = 8 + \int_{2}^{5} f'(x) dx \geq 8 + \int_{2}^{5} (4x-3) dx = 8 + [2x^2-3x]_{2}^{5} = 8 + (50-15) - (8-6) = 8 + 35 - 2 = 41$.Wait,checking the options,$f(5)+f'(5) \geq 11+17 = 28$ is the correct inequality derived from the lower bounds.

Let $f$ be a twice differentiable function on $(1,6)$. If $f(2)=8$,$f'(2)=5$,$f'(x) \geq 1$ and $f''(x) \geq 4$ for all $x \in (1,6)$,then:

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