Find all extreme values (if any) of the given function on the given interval: y = [(1 + x^2)]^1/2   for interval (-2,3)This is a precalculus question using maxima and minima

Expert Answers
rcmath eNotes educator| Certified Educator


We will start by finding the first and second derivative of the function



`y''=[(1+x^2)^(1/2)-x^2/(1+x)^(1/2)]/(1+x^2)=>` To simplify the top part we have the LCD as (1+x^2)^(1/2), that give us 


`y''=1/[(1+x^2)^(1/2)*(1+x^2)]=1/(1+x^2)^(3/2)` Hence y'' is always positive.

To look for the relative max or min we set y'=0=>x=0 and since y'' is positive then we have a local min.

Let's check if it is absolute min on the given interval




Thus (0,1) is an absolute minimum. The following graph confirm our findings.