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