Let f(x) = ln(x^(4)+ 27). Find the intervals of concavity and the inflection pointsof the graph of y=f(x).

Answer 1

First, think about x^4 + 27, then the impact of taking logs. We know that the function is even and will be convex around 0, as the minimum of x^4 + 27 is there, and concave to the right and left.

ln(x^4+27)' =

4x^3/(x^4 + 27)

ln(x^4+27)'' =

[12x^2(x^4+27)-16x^6]/(x^4+27)^2 = ( - 4x^6+324x^2)/(x^4+27)^2 =

4x^2(81-x^4)/(x^4+27)^2

This equals 0 at x = 0, -3, and 3.

As we already mentioned, 0 is the minimum, not an inflection point. The denominator is always positive. Clearly, 81 - x^4 is positive on (-3, 3) and negative on (-∞, -3) and (3, ∞)

Thus, the inflection points are at ±3 and the function is convex on (-3, 3) and concave on (-∞, -3) and (3, ∞)

The global minimum is 3 ln 3 at 0 and there is no global maximum.

The inflection points are (-3, 2 ln 2 + 3 ln 3) and (3, 2 ln 2 + 3 ln 3)