Concavity and inflection points

A graph can rise while bending upwards, or rise while bending downwards. We can distinguish between the cases by where the graph lies relative to its tangent lines. We call this concavity.

We cannot test the graph against every one of its tangent lines directly, so we work from $f''$ instead. Since $f''$ is the derivative of $f'$, its sign tells us whether $f'$ is rising or falling, just as the sign of $f'$ tells us whether $f$ is rising or falling.

Where the graph is concave up, the tangent slope increases from left to right, moving from negative through zero to positive, so $f'$ is rising and $f'' > 0$. Where it is concave down, the slope decreases from left to right, so $f'$ is falling and $f'' < 0$.

We find the concave-up and concave-down intervals the same way we found where $f$ is increasing or decreasing: test the sign of the derivative on each interval between the points where it is zero. Only now the derivative is $f''$.

When $f''$ is linear, each of these is a linear inequality and each solution is a single unbounded interval. To name the concavity at one particular value of $x$, we evaluate $f''$ there and read its sign.

Find where $f(x) = x^3 - 6x^2 + 5x$ is concave up and where it is concave down.

Solution

1. We differentiate $f$ to get $f'$, then differentiate $f'$ to get $f''$.

   $$\begin{align*} f'(x) &= 3x^2 - 12x + 5 \\ f''(x) &= 6x - 12 \end{align*}$$

2. We solve $f'' > 0$ for the values of $x$ where $f$ is concave up.

   $$\begin{align*} 6x - 12 &> 0 \\ 6x &> 12 \\ x &> 2 \end{align*}$$

   So $f$ is concave up on $(2, \infty)$.

3. We solve $f'' < 0$ for the values of $x$ where $f$ is concave down. The same rearrangement with the inequality reversed gives $x < 2$, so $f$ is concave down on $(-\infty, 2)$.

The graph below shows the change of concavity at $x = 2$, from concave down to concave up.

We have found the intervals where a curve is concave up and where it is concave down. The point that separates them, where the concavity changes, is an inflection point.

At such a point the tangent line crosses the curve: the graph lies on one side of the tangent just before the point and on the other side just after.

Since $f$ is concave up where $f'' > 0$ and concave down where $f'' < 0$, an inflection point is a point where $f''$ changes sign. To change sign, $f''$ must pass through zero, so $f''(c) = 0$ at every inflection point $c$.

Solving $f''(x) = 0$ gives the candidates, but $f''(c) = 0$ on its own is not enough: $f''$ can reach zero and keep the same sign on both sides, so the concavity does not change and the candidate is not an inflection point. As with any sign analysis, we confirm each candidate by checking the sign of $f''$ on both sides, and only a genuine change of sign marks an inflection point.

Find the inflection points of $f(x) = x^3 - 3x^2 + 5x$.

Solution

1. We differentiate $f$ to get $f'$, then differentiate $f'$ to get $f''$.

   $$\begin{align*} f'(x) &= 3x^2 - 6x + 5 \\ f''(x) &= 6x - 6 \end{align*}$$

2. We solve $f''(x) = 0$ to find the candidates.

   $$\begin{align*} 6x - 6 &= 0 \\ x &= 1 \end{align*}$$

   So $x = 1$ is the only candidate.

3. We evaluate $f''$ at a value on each side of $x = 1$, taking $x = 0$ and $x = 2$. Here $f''$ is linear, so it keeps one sign on each side of the candidate and any test value in the interval gives that sign.

   $$\begin{align*} f''(0) &= 6(0) - 6 = -6 \\ f''(2) &= 6(2) - 6 = 6 \end{align*}$$

4. The signs differ, so the concavity changes at $x = 1$: concave down to the left, concave up to the right. The candidate is an inflection point. Its height is $f(1) = 1 - 3 + 5 = 3$, so the inflection point is $(1, 3)$.