Question

Is `f(x)=(x-3)^3-x+15` concave or convex at `x=3`?

Answer 1

Neither. `x= 3` is an inflection point.

Explanation:

You have to find the second derivative to answer this question.

I would use the chain rule to differentiate `(x- 3)^3`. Let `u = x - 3` and `y = u^3`. Then `y' = 3u^2 * u' = 3u^2 * 1 = 3(x - 3)^2`

`f'(x) = 3(x- 3)^2 - 1`

Now it's easy enough to expand to find the second derivative using the power rule.

`f'(x) = 3(x^2 - 6x - 9) - 1`

`f'(x) = 3x^2 - 18x - 27 - 1`

`f'(x) = 3x^2 - 18x - 28`

Now find the second derivative.

`f''(x) = 6x - 18`

We now test the sign of the second derivative at `x = 3` under the knowledge that if:

`•f''(x) > 0`, at `x = a` then `f(x)` is concave at `x = a`

`•f''(x) < 0`, at `x = a` then `f(x)` is convex at `x = a`

We have:

`f''(3) = 6(3) - 18 = 18 - 18 = 0`

So `x = 3` is a point of inflection, so the graph is neither concave up nor concave down at that point. Instead its changing between the two.

Hopefully this helps!