Finding and classifying critical numbers on a closed interval

First differentiate: $f'(x) = 1 - \dfrac{4}{x^2}$. Setting $f'(x) = 0$ gives $x^2 = 4$, so $x = \pm 2$. Since $f'(x)$ is undefined only at $x=0$, which is not in $[1,6]$, the only critical number in the interval is $x = 2$ (note $x=-2$ is outside $[1,6]$).

Now evaluate $f$ at the critical number and the endpoints: $f(1) = 1 + 4 = 5$, $f(2) = 2 + 2 = 4$, and $f(6) = 6 + \dfrac{4}{6} = \dfrac{20}{3} \approx 6.67$. Comparing these values, the smallest is $f(2)=4$ and the largest is $f(6) = \dfrac{20}{3}$.

Therefore, the critical number in $[1,6]$ is $x=2$, the absolute minimum value is $f(2) = 4$, and the absolute maximum value is $f(6) = \dfrac{20}{3}$.

The left endpoint is x equals one, where f equals five. The critical number is x equals two, where f equals four. Comparing these two values, five is larger than four. So the left endpoint gives the larger function value.