# 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) #?

##### 1 Answer

If

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

minimum at

#### Explanation:

We have

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) #

# :. y^2 = x^2+1 #

# :. 2ydy/dx = 2x #

If

So s there is just a single critical point when

First note that as

# 2ydy/dx = 2x => dy/dx = x/y #

# :. dy/dx = x/sqrt(x^2+1) #

So we can see immediately 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

Which concludes the deduction