f(x)=ln(x^4+27)

Question

find the inflection points?

Answer 1

f'(x) = 4x^3/(x^4 + 27)

f''(x) = (12x^2(x^4+27) - 4x^3(4x^3))/(x^4+27)^2
= 4x^2(81-x^4)/(x^4+27)^2 = 0 for inflection points

4x^2 = 0 or 81-x^4 = 0
x = 0 or (9-x^2)(9+x)^2 = 0
x = 0 or (3+x)(3-x)(9+x)^2 = 0

x = 0 or x = -3 or x = 3

So we know the x values of our 3 points of inflection, sub into the original equation to find the matching y values.

Answer 2

The inflection points are where the second derivative is zero.

f'(x) = 3x^2/(x^4 +27)
f''(x) = [(x^4+27)(6x) - 12x^5]/(x^4 +27)^2
f''(x) = 0 where the denominator is zero.

Answer 3

i am not getting the ans i mean it says you have to find smaller and larger values.

i already know that one smaller is -3
and one larger value is 3 but the other two i am not getting :(