# 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

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### 1 Answer

`y=f(x)=(1+x^2)^(1/2)`

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

`y'=1/2(1+x^2)^(1/2-1)*2x=(1+x^2)^(-1/2)*x=x/(1+x^2)^(1/2)`

`y''=[1*(1+x^2)^(1/2)-x*1/2*(1+x^2)^(-1/2)*2x]/((1+x^2)^(1/2))^2=>`

`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+x^2)-x^2]/(1+x^2)^(1/2)]/(1+x^2)`

`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

`f(-2)=(1+(-2)^2)^(1/2)=5^(1/2)=sqrt(5)~~2.24`

`f(0)=(1+0)^(1/2)=1`

`f(3)=(1+3^2)^(1/2)=10^(1/2)=sqrt(10)~~3.16`

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