How do you find points of inflection and determine the intervals of concavity given #y=2ln(x^2-1)#?

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Answer 1 · 2017-07-26T13:35:44

Please see below.

Let's call the function #F#.

#f(x) = y = 2ln(x^2-1)#

The domain is #(-oo,-1)uu(1,oo)#.

(The natural logarithm is not defined at #0# and is not real for negatives.)

#f'(x) = (4x)/(x^2-1)#

#f''(x) = (4(x^2-1) - (4x)(2x))/(x^2-1)^2#

# = (-4(x^2+1))/(x^2-1)^2#

Both #x^2+1# and #(x^2-1)^2# are always positive, so #f''(x)# is negative for all #x# in the domain of #f#.

There are no points of inflection and the graph is concave downward (also called concave) on #(-oo,-1)# and on #(1,oo)#.

Note the graph is not concave downward on #(-oo,oo)# because it is not defined on #[-1,1]#.

The graph is not concave downward on its domain. The domain is not an interval. It is the union of two disjoint intervals.

Here is the graph. (You can scroll in/out and drag the view around. When you leave the page and return, you'll see the default original view again.)

graph{2ln(x^2-1) [-16.02, 16.02, -8.01, 8.01]}