What are the local extrema of #f(x) = 2 x + 3 /x#?

Question

Paisley Johnson · Accepted Answer

The local extrema are #-2sqrt(6)# at #x = -sqrt(3/2)#
and #2sqrt(6)# at #x = sqrt(3/2)# Local extrema are located at points where the first derivative of a function evaluate to #0#. Thus, to find them, we will first find the derivative #f'(x)# and then solve for #f'(x) = 0#. #f'(x) = d/dx(2x+3/x) = (d/dx2x) + d/dx(3/x) = 2 - 3/x^2# Next, solving for #f'(x) = 0# #2-3/x^2 = 0# #=> x^2 = 3/2# #=> x = +-sqrt(3/2)# Thus, evaluating the original function at those points, we get #-2sqrt(6)# as a local maximum at #x = -sqrt(3/2)# and #2sqrt(6)# as a local minimum at #x = sqrt(3/2)#

Adam Adkins · Answer

undefined To find the local extrema of $ f(x) = \frac{2x + 3}{x} $, we first find the derivative of $ f(x) $ and then solve for critical points.

$ f'(x) = \frac{d}{dx}\left(\frac{2x + 3}{x}\right) = \frac{(x)(2) - (2x + 3)(1)}{x^2} $

$ f'(x) = \frac{2x - 2x - 3}{x^2} = \frac{-3}{x^2} $

To find critical points, set the derivative equal to zero:

$ \frac{-3}{x^2} = 0 $

This equation has no solutions since the numerator is a constant and cannot be zero. Thus, there are no critical points.

However, $ f(x) $ is not defined at $ x = 0 $, so we need to check the behavior of $ f(x) $ around this point.

As $ x $ approaches $ 0 $ from the right ($ x > 0 $), $ f(x) $ approaches $ -\infty $.

As $ x $ approaches $ 0 $ from the left ($ x < 0 $), $ f(x) $ approaches $ +\infty $.

Therefore, there is a vertical asymptote at $ x = 0 $, but no local extrema for $ f(x) $.