Let y= x + 1/x ,x not =0. Find the intervals on which f is an increasing function?

2 Answers
Aldrich
Sep 9, 2015

#y = x+1/x# is increasing in the intervals #(-oo,-1) uu (1,+oo)#

Explanation:

We're looking for the intervals where #dy/dx > 0#.
Recall that a function is increasing at a point when it's derivative at that point is greater than zero.

Getting the derivative
First, we get the derivative of #x+1/x#.

#d/dx (x+1/x)#

#=d/dxx + d/dx(x^-1)#

#=(1) + (-1)(x^-2)#

#=1-1/(x^2)#

#dy/dx = 1-1/x^2#

Now, we need to solve for:

#1-1/(x^2) > 0#

#1> 1/(x^2) #

Solving for the intervals (one approach)

Since #x^2# is always positive for any real number #x#, we can multiply both sides by #x^2# without changing the sign.

#x^2> 1 #

#x^2 - 1> 0#

The zeroes of #x^2-1# are #{1,-1}#, and the parabola curves upward, since the lead coefficient is greater than #0#.

This results in the graph:

graph{x^2-1 [-2.624, 2.85, -1.727, 1.01]}

Therefore #dy/dx > 0# over the intervals #(-oo,-1) uu (1,+oo)#.

So, the equation is increasing over those intervals.

bp
Sep 10, 2015

The answer can be put into a compact form

Explanation:

Can be put into a compact form as below:

enter image source here

Related questions
Impact of this question
998 views around the world
You can reuse this answer
Creative Commons License