Find the points of local maxima or local minima, if any, of the follow

1.	Find the points of local maxima or local minima, if any, of the following functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be :f(x) = \(\frac{x}{2}\)+\(\frac{2}{x}\), x > 0
Answer» 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

Find the points of local maxima or local minima, if any, of the following functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be :f(x) = \(\frac{x}{2}\)+\(\frac{2}{x}\), x > 0

Answer»

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