We have,
f(x) = \(\frac{x}{2}\)+\(\frac{2}{x}\), x > 0
Differentiate w.r.t x, we get,
f'(x) = \(\frac{1}{2}\)+\(\frac{2}{x^2}\), x > 0
For the point of local maxima and minima,
f’(x) = 0
\(\frac{1}{2}\) - \(\frac{2}{x^2}\) = 0
= x2 – 4 = 0
= x = \(\sqrt4\) , \(-\sqrt4\)
= x = 2, – 2
At x = 2 f’(x) changes from –ve to + ve
Since, x = 2 is a point of Minima
Hence, local min value f (2) = 2