Points of Inflection Examples

Find all points of inflection for the function f (x) = x³.

We want to find where the second derivative changes sign, so first we need to find the second derivative. The first derivative is

f '(x) = 3x²

and the second derivative is

f "(x) = 6x.

The second derivative is never undefined, and the only root of the second derivative is x = 0. The sign of f " does change there, since f "(x) is negative for x < 0 and positive for x > 0.

Therefore x = 0 is an inflection point. Looking at the graph of f (x), this makes sense:

Find all inflection points for the function f (x) = x⁴.

The first derivative is f '(x) = 4x³ and the second derivative is

f "(x) = 12x².

The second derivative is never undefined, and the only root of the second derivative is x = 0. However, f "(x) is positive on both sides of x = 0, so the concavity of f is the same to the left and to the right of x = 0. This means x = 0 is not a point of inflection.

Since x = 0 was the only root of f "(x), the function f has no inflection points.
Looking at the graph of f (x), this makes sense:

Suppose we started with a function f, took a couple of derivatives, and found that

f "(x) = (x + 2)(x + 1).

What are the inflection points of f ?

f " is never undefined. The roots of f " are -2 and -1:

Figure out what the sign of f " is doing. 

Way 1: The roots of f " break up the numberline into intervals:

Take a number from each interval, plug it into f ", and see if you find a positive or a negative value:

If you find a positive value, f " will be positive for that whole interval. If you find a negative value, f " will be negative for that whole interval:

Way 2: Way 2 is much like Way 1, except that we don't bother to plug in actual numbers. We still need to look at each interval, though:

When x < -2, the quantity (x + 2) will be negative and the quantity (x + 1) will also be negative. Therefore f "(x) will be the product of two negative numbers, and will be positive.

When -2 < x < -1, the quantity (x + 2) will be positive. This is because if x is closer to zero than -2, adding 2 to x will bump it up into the positive numbers. However, (x + 1) will still be negative, since adding 1 to x will not be enough to bump it into the positive numbers. Therefore f "(x) will be the product of one negative and one positive number, therefore negative.

When -1 < x, both (x + 2) and (x + 1) will be positive, so f "(x) will be positive.

Thankfully, this gives us the same picture we had before:

Since the sign of f " does change at both x = -2 and x = -1, both of these would be inflection points of f.