AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Offline Centres

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values as the case may be: $g (x) = \frac{x}{2} + \frac{2}{x}, x > 0$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

g (x) = \frac{x}{2} + \frac{2}{x}

g ` (x) = \frac{1}{2} - \frac{2}{x^{2}}

Put

g ` (x) = 0

\Rightarrow \frac{1}{2} - \frac{1}{x^{2}} = 0

\Rightarrow x = \pm 2

Since x is given as greater than 0, so we get the value of x as 2 only.

g ` ` (x) = \frac{4}{x^{3}}

g ` ` (2) = \frac{4}{2^{3}} = \frac{1}{2} > 0

Therefore 2 is the point of local minima by second derivative test.
Putting x = 2 in g(x) we get,

g (2) = \frac{2}{2} + \frac{2}{2} = 2

Hence, the local minimum value is 2.

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test