Determining Extrema Determining Extrema

The First Derivative Test

The First Derivative Test states, for a continuous function

f
, that if
f'(x)
changes sign at
x=c
, then
f(x)
has a local minimum or maximum at
x=c
.

When
f'
crosses the x-axis going from positive to negative,
f
has a local maximum. When
f'
crosses the x-axis going from negative to positive,
f
has a local minimum.
In this graph, the derivative does not change sign.
f
does not have a local minimum or maximum.

The Candidates Test

The Candidates Test states that, for a continuous function

f
on a closed interval
[a,b]
, the global minimum and maximum of
f
can only occur at
a
,
b
, or the critical points of
f
.

The global minimum and maximum of
f(x)=x^3 - x
on the interval
[-1,3]
are at
(\frac{1}{\sqrt{3}},\frac{-2}{3\sqrt{3}})
and
(3,24)
, respectively.

The Second Derivative Test

For a continuous function,

f
, if
f'(c)=0
and
f''(c) \gt 0
, then
f
has a local minimum at
x=c
. If
f'(c)=0
and
f''(c) \lt 0
, then
f
has a local maximum at
x=c
.

In other words, local minimums occur when the function is concave up (the slope is increasing), and local maximums occur when the function is concave down (the slope is decreasing).

Here is the same graph with the second derivative, to see how the sign of the second derivative matches the concavity of the function.

One Critical Point Test

For a continuous function,

f
, on a closed interval,
[a,b]
, if
f
has only one critical point, and that critical point is a local minimum/maximum, then that critical point is a global minimum/maximum.

If a function has only one critical point, then its slope can only change sign once. In other words, there is only one spot where the function could "turn around". If the function does turn around there, then it is a global min/max.

The function
f(x)=x^2
on the interval
[-3,3]
has only one critical point,
(0,0)
, which is a local minimum. Therefore, it is also the global minimum.