# Inflection Points

## Related Calculator: Inflection Points and Concavity Calculator

**Definition**. Point `c` is an **inflection point** of function `y=f(x)` if function at this point changes direction of concavity (i.e. from concave upward becomes concave downward or from concave downward becomes concave upward).

It appears that `f''(x)` plays same role in finding inflection points as played `f'(x)` in finding extrema.

**Fact**. If point `c` is inflection point of the function `y=f(x)` then either `f''(c)=0` or `f''(c)` doesn't exist.

Converse is not true. Consider function `f(x)=x^4`. `f''(x)=12x^2`, so `f''(0)=0`, but `x=0` is not an inflection point.

But we know that when `f''(x)>0` function is concave upward, and when `f''(x)<0` function is concave downward.

So, point will be point of inflection if second derivative changes sign at this point.

This rule is very similar to First Derivative Test for extrema.

**Procedure for Finding Inflection Points**:

- Find points where `f''(x)=0` or `f''(x)` doesn't exist.
- Test all points in the following way: if second derivative changes sign at point then it is point of inflection. Otherwise, it is not a point of inflection.

Now it is clear why `x=0` is not point of inflection for `f(x)=x^4`. That's because `f''(x)=12x^2>=0` and second derivative doesn't change sign at `x=0`.

**Example 1 **. Find points of inflection of `f(x)=x^(1/3)`.

We have that `f'(x)=1/3 x^(-2/3)` and `f''(x)=-2/9x^(-5/3)`.

Second derivative is never zero, but it doesn't exist at `x=0`.

Since `f''(x)>0` for `x<0` and `f''(x)<0` for `x>0` then second derivative changes sign at `x=0`.

This means that `x=0` is point of inflection.

**Example 2**. Find points of inflection of `f(x)=sin(x)`.

Since `f'(x)=-cos(x)` and `f''(x)=-sin(x)`. then points of inflection are points where `sin(x)=0` (clearly at these points second derivative changes sign).

Thus, points of the form `x=pik,k in ZZ` are points of inflection.

This means that `sin(x)` has infinite number of points of inflection.

**Example 3**. Find points of inflection of `f(x)=3x^5-5x^4+5x-7`.

We have that `f'(x)=15x^4-20x^3+5` and `f''(x)=60x^3-60x^2=60x^2(x-1)`.

Derivative is defined for all `x` so there are no such points where derivative doesn't exist.

So, inflection points are among roots of equation `f''(x)=0`.

Equation `60x^2(x-1)=0` gives two roots: `x=0` and `x=1`.

Now use method of intervals to where second derivative is positive and where negative.

Number line is divided by points `x=0` and `x=1` on 3 intervals: `(-oo,0)uu(0,1)uu(1,oo)`.

First interval is `(-oo,0)`: here `x^2>0` and `x-1<0` so `f''(x)<0`.

Second interval is `(0,1)`: here `x^2>0` and `x-1<0` so `f''(x)<0`.

First interval is `(1,oo)`: here `x^2>0` and `x-1>0` so `f''(x)>0`.

We see that derivative doesn't change sign at `x=0` and changes sign from negative to positive at `x=1`.

This means that the only point of inflection is `x=1`.

As in case with extrema we can formulate criterion of finding inflection points using higher-order derivatives.

**Fact**. Suppose that for function `y=f(x)` there is point `c` such that `f''(c)=0`. Also suppose `k` (`k>2`) is the smallest number such that `f^((k))(c)!=0`. If `k` is odd then `c` is point of inflection, if `k` is even then there is no point of inflection.

Let's solve Example 3 using above fact.

We have that `f'''(x)=180x^2-120x`.

Since `f'''(1)=60!=0` and 3 is odd number, then `x=1` is point of inflection.

Since `f'''(0)=0`, then third derivative is inconclusive for `x=0`.

Find fourth derivative `f^((4))(x)=360x-120`.

Since `f^((4))(0)=-120!=0` and 4 is even number, then `x=0` is not a point of inflection.