Question

Are the inflection points where f'(x) = zero or where the graph changes from concave up to concave down?

Answer 1

The inflection point is a point where the graph of the function changes from concave up to concave down or vice versa.

To calculate these points you have to find places where `f''(x)=0` and check if the second derivative changes sign at this point.

For example to find the points of inflection for `f(x)=x^7`you have to calculate `f''(x)` first.

`f'(x)=7x^6`
`f''(x)=42x^5`

Now we have to check where `f''(x)=0`
`42x^5=0 iff x=0`.

We found that `x=0` may be a point of inflection.
To find if it is such point we have to check if `f''(x)` changes sign at 0.

To find this we can graph the function:

graph{42x^5 [-3.894, 3.897, -1.95, 1.948]}

We can see that the `f''(x)` changes sign at zero, so zero is the point of inflection.

Note
It is important to check to see whether concavity actually changes.

`g(x)=x^4+3x-8`

`g'(x)+ 4x^3+3`

`g''(x) =12x^2`

Now we have to check where `g''(x)=0`
`12x^2=0 iff x=0`.

We found that `x=0` may be a point of inflection.
To find if it is such point we have to check if `g''(x)` changes sign at `0`.

But `g''(x) =12x^2` is never negative, it is always positive or `0`, therefors the sign of `g''(x)` does not change, so there are no inflection points.

Answer 2

I have been taught and, following our textbook's lead, I continue to teach , that an inflection point is a point on the graph at which the concavity changes.

Explanation:

Using this terminology:
For `f(x) = x^7`, the inflection point is `(0,0)`

The function: `h(x) = 1/x` is concave down on `(-oo,0)` and concave up on `(0,oo)`.
The concavity is not the same on the entire graph, but there is no inflection point, because there is no point on the graph at `x=0`.