What are the absolute extrema of  f(x)= x^(2)+2/x  on the interval [1,4]?

We need to find the critical values of $f \left(x\right)$ in the interval $\left[1 , 4\right]$.

Hence we calculate the roots of the first derivative so we have

$\frac{\mathrm{df}}{\mathrm{dx}} = 0 \implies 2 x - \frac{2}{x} ^ 2 = 0 \implies 2 {x}^{2} \left(x - 2\right) = 0 \implies x = 2$

So $f \left(2\right) = 5$

Also we find the values of $f$ at the endpoints hence

$f \left(1\right) = 1 + 2 = 3$

$f \left(4\right) = 16 + \frac{2}{4} = 16.5$

The largest function value is at $x = 4$ hence $f \left(4\right) = 16.5$ is the absolute maximum for $f$ in $\left[1 , 4\right]$

The smallest function value is at $x = 1$ hence $f \left(1\right) = 3$ is the absolute minimum for $f$ in $\left[1 , 4\right]$

The graph of $f$ in $\left[1 , 4\right]$ is