Today, we will be discussing the first derivative test in more detail. We will
begin with a warmup problem.
Example 0.1. Find and classify the critical points of the function
f (x) = x4 + 3x2 + 1.
To do this problem, we differentiate and set the derivative equal to zero:
f ′ (x) = 4x3 + 6x
Clearly, this has a zero at x = 0. After factoring out a copy of x, we have
0 = x(4x2 + 6).
But 4x2 + 6 is always positive, so there are no solutions to 4x2 + 6 = 0. Therefore
0 is the only critical point of f (x).
We should plug in a negative number and a positive number into f ′ (x) to determine what kind of critical point this is. We can calculate that f ′ (−1) = −10 and
f ′ (1) = 10, so we get that x = 0 is a local minimum for this function.
The main takeaway from this problem is that not every factor of the derivative
will contribute a zero. Sometimes even a high-degree polynomial will have only
one critical point. Don’t overthink the algebra here- if a factor of the derivative is
obviously nonzero, that’s not a problem.
Example 0.2. Find and classify the critical points of
x2 − 1
.
x+3
We can differentiate this by using the quotient rule:
f (x) =
f ′ (x) =
(x + 3)(2x) − (x2 − 1)
(x + 3)2
which can be rewritten as
x2 + 6x + 1
.
(x + 3)2
So f (x) is not differentiable at x = −3. As it turns out, f (x) isn’t even defined at
x = −3.
We only need to see where the numerator is zero:
f ′ (x) =
x2 + 6x + 1 = 0
Using the quadratic formula, we get
√
−6 ± 32
2
so
√
x = −3 ± 8
√
We know that 8 is between 2 and 3. So let’s test f ′ :
Notice that the denominator of f ′ (x) is always positive so it’s only necessary to
check the sign of the numerator. We plug in −6 and get that the numerator is 1,
which is positive. Plugging in −4 gives that the numerator is −7. If we plug in −2,
we also get 07, and
√ if we plug in x = 0, then we get
√ 1.
So x = −3 − 8 is a maximum and x = −3 + 8 is a minimum.
Note that we needed to check on both sides of the point x = −3; it’s possible
for the derivative to change sign there.
x=
1
2
Let’s do an example that isn’t a rational function:
Example 0.3. Find and classify the critical points of
1
f (x) = e2x − ex .
2
We differentiate to get
f ′ (x) = e2x − ex
′
Now, we set f (x) = 0 and factor out a copy of ex :
ex (ex − 1) = 0
For the left hand side we need either ex = 0 or ex − 1 = 0. But ex is always positive,
so it’s never zero. So we only need to consider the case for which ex − 1 = 0, which
happens when x = 0.
We should be able to guess what kind of critical point this is in advance, but it’s
not hard to see that it’s a minimum- at x = −1, the derivative is e−2 − e−1 , which
is negative, and at x = 1, the derivative is e2 − e1 , which is positive.