Let f(x)=ln( x^2 + 1). Determine the open intervals on which f is concave up and concave down?

Question

Also, find the coordinates of the point(s) of inflection of f. Specify both the x and y coordinates of the point(s) of inflection. Give answers in exact form.

Please Help!

Answer 1

f(x) = ln( x^2 + 1)
f'(x) = 2x / ( x^2 + 1)
f'(x) = - 4x^2( x^2 + 1)^-2 + 2( x^2 + 1)^-1
At points of inflection,
4x^2( x^2 + 1)^-2 = 2( x^2 + 1)^-1
2x^2 = x^2 + 1
x^2 = 1
x = ± 1
y = ln(2)

x,y = (-1,ln(2)), (1, ln(2))
The function is concave down for
x < - 1 and x > 1
It is concave up for
-1 < x < 1

Answer 2

Here, you want to find the second derivatives of the function and determine the intervals in which that second derivative is positive or negative. Positive 2nd deriv. implies concave up, and negative 2nd deriv. implies concave down.
From the chain rule,
f'(x) = (1/(x^2 +1))*(2x) = 2x/(1+x^2)
f''(x) =[ 2*(1+x^2) - 4x^2 ] / (1+x^2)^2

Then denominator is positive for all real x, so when the numerator is zero, you have the coordinates of your inflection points.
Set to zero,
2*(1+x^2) - 4x^2 =0
2+2x^2 = 4x^2
1 = x^2 at inflection points.
Therefore, inflection points occur at x= +/- 1
The numerator is negative, and thus the function is concave down, when
1-x^2 <0
when |x| is greater than 1,
and is concave up when -1

Answer 3

hi

y = ln (x^2 + 1)

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

y'' = [ 2 * (x^2 + 1) - 2x * 2x] / (x^2 + 1)^2

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

inflection x = +/- 1 -> y = ln 2

for x < -1 or 1 < x
y" < 0 concave down

for -1 < x < 1
y" > 0 concave up

bye

Answer 4

yeah one and two answered it