Where is this function decreasing?

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2 Answers
Jan 9, 2018

(#color(red)(-1)#,#color(blue)(" 1")#) #(1, oo)#

Explanation:

This function is decreasing when the y-value decreases.

In interval notation this is written as so:
Dec(#color(red)(-1)#,#color(blue)(" 1")#) #(1, oo)#
The #color(red)"red"# number is the x-value that the decreasing interval starts and the #color(blue)"blue"# number is the x-value that the decreasing interval ends.

The function also decreases at the end as x approaches positive infinity.

Jan 9, 2018

This function is decreasing in the intervals #(0, 1)# and #(1, oo)#

Explanation:

A function #f(x)# is decreasing at a point #x=a# if there is some #epsilon > 0# such that both of the following hold:

#f(x) > f(a)# for all #x in (a-epsilon, a)#

#f(x) < f(a)# for all #x in (a, a+epsilon)#

If the function has a well defined tangent at the point #x=a# then the slope of the tangent will be negative.

In the given example, note that for any #x in (0, 1) uu (1, oo)#, there is a small neighbourhood of #x# such that the function is greater to the left and lesser to the right. So the function is decreasing in this union of intervals.

Bonus

Given that the function has vertical asymptotes at #x=+-1#, horizontal asymptote #y=0# and #y# intercept #(0, -2)#, we can make a guess at an equation for the function:

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

graph{2/(x^2-1) [-10, 10, -12, 12]}