5.5 Monotonicity and Extremum

1. Increasing - Decreasing Functions

Consider the following graph of $y = f(x)$:

a c₁ c₂ c₃ c₄ c₅ b x y y = f(x)

Let us make some observations from the graph:

  • The domain of the graph is the interval $[a, b]$. The points $x=a$ and $x=b$ are called endpoints.
  • We state that we have a local max (or just max) at the points: $x=c_1, x=c_3, x=b$.
  • We state that we have a local min (or just min) at the points: $x=a, x=c_2, x=c_4$.
All these points (max and min) are called turning points or extreme values.
Notice that $x=c_5$ is not a turning point (neither max nor min), as near $f(c_5)$ you can find smaller as well as larger values.

Tracking the flow of the curve:

  • The function is increasing (goes up) in the interval $(a, c_1)$
  • The function is decreasing (goes down) in the interval $(c_1, c_2)$
  • The function is increasing (goes up) in the interval $(c_2, c_3)$
  • The function is decreasing (goes down) in the interval $(c_3, c_4)$
  • The function is increasing (goes up) in the interval $(c_4, b)$

Remember that a positive gradient means that the function is increasing (goes up), and a negative gradient implies the function is decreasing (goes down). But we know that derivative = gradient.

In other words:
If $f'(x) > 0$ then $f$ is increasing ($\nearrow$)
If $f'(x) < 0$ then $f$ is decreasing ($\searrow$)

Notice: The increasing or decreasing behavior of a function is known as its monotonicity!

2. Turning Points: MAX - MIN

How can we isolate the turning points (max or min) of a function?

First, the endpoints are extreme values (see $a$ and $b$ above). As far as the interior points are concerned, observe that the gradient at any turning point is 0 (the tangent lines at those values are horizontal!).

PROPOSITION

If $f(x)$ has a turning point (max or min) at some interior point $c$ and $f'(c)$ exists, then:

$f'(c) = 0$

Notice that $f'(x) = 0$ at $c_1, c_2, c_3, c_4$, and $c_5$. We verify a local max at $c_1, c_3$ and a local min at $c_2, c_4$. However, $c_5$ is not a turning point. Hence, the inverse proposition is not true.

Therefore, apart from the endpoints, the possible turning points (max/min) are restricted to the following:

  • points $x$ where $f'(x) = 0$ (called stationary points)
  • points where $f'(x)$ does not exist

All these boundaries are classified as critical points. Here we deal only with stationary points (where $f'(x) = 0$). To verify whether a stationary point $x=c$ acts as a turning point (max or min), we execute the following test.

FIRST DERIVATIVE TEST for c

Check the sign of $f'(x)$ to verify if the curve is increasing or decreasing just before and after $c$:

$x$ $c$
$f'(x)$ + 0 $-$
Conclusion for $f$ $\nearrow$ max $\searrow$
$x$ $c$
$f'(x)$ $-$ 0 +
Conclusion for $f$ $\searrow$ min $\nearrow$

If the sign does not alter, we possess neither a max nor a min.

METHODOLOGY

Given $y = f(x)$:

  • Step 1: we find $f'(x)$
  • Step 2: we solve $f'(x) = 0$ (say the resulting roots are $a, b, c$)
  • Step 3: we construct a table as follows to perform the first derivative test:
$x$ $a$ $b$ $c$
$f'(x)$ + 0 $-$ 0 + 0 +
Conclusion for $f$ $\nearrow$ max $\searrow$ min $\nearrow$ nothing $\nearrow$

EXAMPLE 1

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

We find $f'(x) = x^2 - 4x + 3$
We solve $x^2 - 4x + 3 = 0$
The solutions are $x=1$ and $x=3$
We construct the table:
$x$ 1 3
$f'(x) = x^2 - 4x + 3$ + 0 $-$ 0 +
Conclusion for $f$ $\nearrow$ max $\searrow$ min $\nearrow$
Therefore, we evaluate:
we have a max at $x=1$ [and the max value of $f$ is $f(1) = 6.33$]
we have a min at $x=3$ [and the min value of $f$ is $f(3) = 5$]

An alternative mechanism to verify if a stationary point correlates to a max or a min is the following:

SECOND DERIVATIVE TEST for c

Find $f''(x)$ (if it exists!)

  • If $f''(c) > 0$ then $c$ establishes a min
  • If $f''(c) < 0$ then $c$ establishes a max
  • If $f''(c) = 0$ we do not attain a conclusive answer. We revert back to the first derivative test.

EXAMPLE 2

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

We established $f'(x) = x^2 - 4x + 3$ and isolated the stationary points $x=1$ and $x=3$.
We differentiate to find $f''(x) = 2x - 4$.

For $x=1$, $f''(1) = -2 < 0$, so we confirm a max at $x=1$.
For $x=3$, $f''(3) = 2 > 0$, so we confirm a min at $x=3$.

EXAMPLE 3

Consider $f(x) = (x-1)^4$

We find $f'(x) = 4(x-1)^3$.
There is one stationary point at $x=1$.
Evaluating $f''(x) = 12(x-1)^2$.
For $x=1$, $f''(1) = 0$ (neither positive nor negative). Thus, we cannot conclude if it represents a max or a min.
The table of signs gives:
$x$ 1
$f'(x) = 4(x-1)^3$ $-$ 0 +
Conclusion for $f(x)$ $\searrow$ min $\nearrow$
Therefore, we identify a min at $x=1$.