5.6 Concavity and Points of Inflection

1. Concavity

Consider the graph of a function from the preceding section:

d₁ d₂ d₃ d₄ d₅ d₆ d₇ x y y = f(x)

Our concern is different! It is to investigate the intervals where the curve:

  • looks like ($\cup$): we say that the function is concave up
  • looks like ($\cap$): we say that the function is concave down
To be formal, a function is strictly concave up/down if the tangent line at each point resides under/above the curve.

We observe that:

  • The function is concave down ($\cap$) in the interval $(d_1, d_2)$
  • The function is concave up ($\cup$) in the interval $(d_2, d_3)$
  • The function is concave down ($\cap$) in the interval $(d_3, d_4)$
  • The function is concave up ($\cup$) in the interval $(d_4, d_5)$
  • The function is concave down ($\cap$) in the interval $(d_5, d_6)$
  • The function is concave up ($\cup$) in the interval $(d_6, d_7)$

The concavity changes at the points $x = d_2, d_3, d_4, d_5, d_6, d_7$.
These boundaries are classified as points of inflexion (or POI).

It is easy to verify the concavity by analyzing the second derivative $f''(x)$:

If $f''(x) > 0$ then $f$ is strictly concave up ($\cup$)
If $f''(x) < 0$ then $f$ is strictly concave down ($\cap$)

Short Explanation (Mainly for HL)

Look at the curve of the following concave up function $f(x)$:

The gradient is negative in the beginning, it is "less" negative as we move forward, it becomes 0, and then becomes positive and "more" positive as we move forward.

In other words, the gradient increases. That means the derivative of the gradient $f'(x)$ is increasing.
We know that the derivative of an increasing function evaluates to positive (+).
Hence, the derivative of $f'(x)$, which is $f''(x)$, is positive (+)!

Similarly, if the function $f(x)$ is concave down, the second derivative must be negative (-)!

2. Points of Inflexion

How can we find the points of inflexion?
Since the concavity changes at such a point, the sign of $f''(x)$ alters from + to - or vice-versa.
Therefore, the second derivative at any valid point of inflexion must evaluate to 0.

PROPOSITION

If $f(x)$ has a point of inflexion at some point $d$ and $f''(d)$ exists, then:

$f''(d) = 0$

Notice that solving the equation $f''(x) = 0$ gives us the possible points of inflexion. To verify if a root $x=d$ is a point of inflexion, we must check the sign of $f''(x)$ before and after that point.

METHODOLOGY

Given $y = f(x)$:

  • Step 1: we find $f'(x)$ and $f''(x)$
  • Step 2: we solve $f''(x) = 0$ (say the roots are $a, b, c$)
  • Step 3: we construct a table of signs to evaluate the boundaries:
$x$ $a$ $b$ $c$
$f''(x)$ + 0 $-$ 0 $-$ 0 +
Conclusion for $f$ $\cup$ poi $\cap$ nothing $\cap$ poi $\cup$

EXAMPLE 1

Consider $f(x) = \dfrac{1}{3}x^3 - 2x^2 + 3x + 5$

We find:
$f'(x) = x^2 - 4x + 3$
$f''(x) = 2x - 4$

We solve $f''(x) = 0 \implies 2x - 4 = 0$. The solution is $x=2$.
We construct the table:
$x$ 2
$f''(x) = 2x - 4$ $-$ 0 +
Conclusion for $f$ $\cap$ poi $\cup$
Therefore, $x=2$ is a point of inflexion.

Let us summarize the information we have derived for $f(x) = \dfrac{1}{3}x^3 - 2x^2 + 3x + 5$ to sketch the graph.

  • First, it helps to evaluate the y-intercept and the x-intercepts (the roots) of $f(x)$ if possible.
  • y-intercept: For $x=0$, $y=f(0)=\mathbf{5}$
  • x-intercepts: We must solve $f(x)=0$ (feasible by GDC as the degree is 3).

We consider a summary table compiling the solutions of both $f'(x)=0$ and $f''(x)=0$:

$x$ 1 2 3
$f'(x)$ + 0 $-$ $-$ $-$ 0 +
$f''(x)$ $-$ $-$ $-$ 0 + + +
Conclusion for $f$ $\nearrow \cap$ max $\searrow \cap$ poi $\searrow \cup$ min $\nearrow \cup$

The table of values grounds the scale:

$x$ 0 1 2 3
$f(x)$ 5 6.33 5.66 5
x y 1 2 3 5 6 max i.p. min

EXAMPLE 2

Consider the exponential function $f(x) = x e^x$.
Evaluate possible maximum, minimum values and points of inflexion.

Solution:
We apply the product rule:
$f'(x) = 1 \cdot e^x + x \cdot e^x = e^x(x+1)$

Stationary points:
$f'(x) = 0 \iff e^x(x+1) = 0 \iff \mathbf{x = -1}$

We use the first derivative test:
$x$ -1
$f'(x)$ $-$ 0 +
Conclusion for $f$ $\searrow$ min $\nearrow$
Then we evaluate the second derivative:
$f''(x) = (x e^x + e^x)' = (e^x + x e^x) + e^x = x e^x + 2e^x$

Points of Inflexion:
$f''(x) = 0 \iff e^x(x+2) = 0 \iff \mathbf{x = -2}$

We construct the table of concavity:
$x$ -2
$f''(x)$ $-$ 0 +
Conclusion for $f$ $\cap$ poi $\cup$

The graph of the function is shown below:

x y -1 -2 -3 1 2
Notice: Look at the curve of this function on your GDC to confirm that $x=-1$ establishes a minimum and observe that at $x=-2$ there is a point of inflexion.