Which of the following?

Question

Which of the following is true about the graph of y=ln abs(x^2-1) in the interval (-1, 1)?

 

a) It is increasing.

b) It attains a relative minimum at (0, 0).

c) It has a range of all real numbers.

d) It is concave down.

e) It has an asymptote of x=0.

 

Answer: D.

Answer 1

In the interval (-1, 1),

y = ln(1-x^2)

y' = -2x/(1-x^2)

y' > 0 in (-1, 0), y is increasing.

y' < 0 in (0, 1), y is decreasing.

y'' = -2[(1-x^2) + 2x^2]/(1-x^2)^2 = -2(1+x^2)/(1-x^2)^2 < 0 in (-1, 1).

So, it is concave down.

 

Answer: D

Answer 2

Hey Sun -- it may help "file away" two (2) key points: ln [0] approaches neg. infinity ...

 

ln [1] = 0 ... thus, the x= +/- 1 maps to "the abyss" and x=0 peaks at y=0 in [-1,1] ...

 

concave down ... Best regards :)

Answer 3

f( x) =

 ln l  X^2 - 1 l  interval ( -1, 1) .

 

     

  X ε ( -1, 0 ) f(x) increasing

   X ε ( 0, 1) f(x) Decreasing.

 

  Graph it, you could see. 

 

 It seems like D is the only correct answer.