Question:

Given that $f$ is a twice differentiable function on $(1,6)$ with $f(2)=8$, $f'(2)=5$, $f'(x)\geq 1$, and $f''(x)\geq 4$ for all $x\in(1,6)$. Step 1: Determine a lower bound for $f'(x)$. Since $f''(x)\geq 4$ for all $x\in(1,6)$, we integrate this inequality from $2$ to $x$: $$ \int_2^x f''(t)\,dt \geq \int_2^x 4\,dt $$ $$ f'(x) - f'(2) \geq 4(x-2) $$ Substitute the given value $f'(2)=5$: $$ f'(x) \geq 5 + 4(x-2) $$ $$ f'(x) \geq 4x-3 $$ Now, we find a lower bound for $f'(5)$ by substituting $x=5$: $$ f'(5) \geq 4(5)-3 = 20-3 = 17 $$ Step 2: Determine a lower bound for $f(x)$. Using the lower bound for $f'(x)$ derived in Step 1, $f'(x) \geq 4x-3$, we integrate this inequality from $2$ to $x$: $$ \int_2^x f'(t)\,dt \geq \int_2^x (4t-3)\,dt $$ $$ f(x) - f(2) \geq \left[2t^2-3t\right]_2^x $$ $$ f(x) - f(2) \geq (2x^2-3x) - (2(2)^2-3(2)) $$ $$ f(x) - f(2) \geq (2x^2-3x) - (8-6) $$ $$ f(x) - f(2) \geq 2x^2-3x-2 $$ Substitute the given value $f(2)=8$: $$ f(x) \geq 8 + 2x^2-3x-2 $$ $$ f(x) \geq 2x^2-3x+6 $$ Now, we find a lower bound for $f(5)$ by substituting $x=5$: $$ f(5) \geq 2(5)^2 - 3(5) + 6 $$ $$ f(5) \geq 2(25) - 15 + 6 $$ $$ f(5) \geq 50 - 15 + 6 $$ $$ f(5) \geq 35 + 6 = 41 $$ Step 3: Combine the lower bounds for $f(5)$ and $f'(5)$. From Step 1, we have $f'(5) \geq 17$. From Step 2, we have $f(5) \geq 41$. Adding these inequalities: $$ f(5) + f'(5) \geq 41 + 17 $$ $$ f(5) + f'(5) \geq 58 $$ Since $f(5)+f'(5) \geq 58$, it follows that $f(5)+f'(5) \geq 28$.