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) #?

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

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.

To determine where the function \( f(x) = \sqrt{x^2+1} \) is increasing or decreasing, and to find the relative maxima and minima, you can follow these steps: 1. Find the derivative of the function \( f(x) \) with respect to \( x \), denoted as \( f'(x) \). 2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points. 3. Determine the sign of \( f'(x) \) in the intervals formed by the critical points. 4. If \( f'(x) > 0 \), the function is increasing in that interval. If \( f'(x) < 0 \), the function is decreasing in that interval. 5. To find relative maxima and minima, examine the behavior of the function around the critical points. A critical point where the function changes from increasing to decreasing is a relative maximum, and a critical point where the function changes from decreasing to increasing is a relative minimum. Now, applying these steps to \( f(x) = \sqrt{x^2+1} \): 1. Find the derivative: \[ f'(x) = \frac{x}{\sqrt{x^2+1}} \] 2. Set \( f'(x) = 0 \): \[ \frac{x}{\sqrt{x^2+1}} = 0 \] \[ x = 0 \] 3. Determine the sign of \( f'(x) \): \[ f'(x) > 0 \text{ when } x < 0 \] \[ f'(x) < 0 \text{ when } x > 0 \] 4. This implies that \( f(x) \) is increasing for \( x < 0 \) and decreasing for \( x > 0 \). 5. Since the function changes from increasing to decreasing at \( x = 0 \), it has a relative maximum at \( x = 0 \). There are no relative minima because the function decreases indefinitely as \( x \) approaches positive or negative infinity.