Question

How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for `f(x) = sqrt(x^2+1)`?

Answer 1

If # { (x<0, => f'(x)<0, =>, "strictly decreasing"), (x=0, => f'(x)=0, =>, "stationary"), (x>0, => f'(x)>0, =>, "strictly increasing") :} #

minimum at `(0,1)`

Explanation:

We have `f(x)=sqrt(x^2+1)`

graph{sqrt(x^2+1) [-10, 10, -5, 5]}

We can deduce from the graph that

f(x) is # { ("strictly decreasing", x<0), ("stationary", x=0), ("strictly increasing", x>0) :} #

So let's prove this using calculus.

For the sake of simpler notation, Let `y= sqrt(x^2+1)`, and we can rearrange and differentiate implicitly. We don't need an explicit expression for `dy/dx` we just need to identify the critical point (`dy/dx=0`)

`y = sqrt(x^2+1)`
`:. y^2 = x^2+1`
`:. 2ydy/dx = 2x`

If `dy/dx=0=>(2y)(0)=2x => x=0 => y=1`

So s there is just a single critical point when `x=0` so we need to look at the sign of `dy/dx` for `x<0` and `x>0`

First note that as `y = sqrt(x^2+1) => y>=1 AA in RR`, as `x^2>=0`

`2ydy/dx = 2x => dy/dx = x/y`
`:. dy/dx = x/sqrt(x^2+1)`

So we can see immediately that `dy/dx` takes the sign of `x`, so we can conclude that:

If # { (x<0, => f'(x)<0, =>, "strictly decreasing"), (x=0, => f'(x)=0, =>, "stationary"), (x>0, => f'(x)>0, =>, "strictly increasing") :} #

And as there is a single stationary point at `(0,1)` this stationary point has to be a minimum.

Which concludes the deduction