Question

What are the points of inflection of #f(x)=12x^3 + 3x^2 + 42x #?

Answer 1

`x=-1/12`

Explanation:

Points of inflection occur when the second derivative a function changes sign (from positive to negative, or vice versa). This correlates to when the concavity of the function shifts.

First, find the second derivative of the function.

`f(x)=12x^3+3x^2+42x`

`f'(x)=36x^2+6x+42`

`f''(x)=72x+6`

The sign of `f''(x)` could shift when `f''(x)=0`, so set `72x+6=0`.

`72x+6=0`

`x=-6/72`

`x=-1/12`

Analyze the sign of `f''(x)` around the possible point of inflection `x=-1/12`.

When `mathbf(x<-1/12)`:

`f''(-1)=-72+6=-66`

This is `<0`.

When `mathbf(x> -1/12)`:

`f''(0)=6`

This is `>0`.

Thus, the sign of `f''(x)` does change around the point when `x=-1/12`.

We can check a graph of `f(x)`:

graph{12x^3+3x^2+42x [-2.5, 2.5, -200, 200]}

The concavity does seem to shift very close to `x=0`.