5.2 - Relative Extrema, Graphing Polynomials
The point (x ,0(x )0 is considered a relative max if there is an open interval containing x on 0
which f(x ) is the largest value. In other words, graphically, it’s the highest point in a general
area. It doesn’t have to be the strictly highest point in total on the graph (although, that is also a
relative max), but just a highest point in some surrounding area. Think mountains. You got a
whole bunch of mountain peaks. Yes, there is probably only one tallest peak, but there are
certainly a whole bunch more of smaller peaks. The relative max of a function would be all those
peaks in the mountains.
Just like how there’s a concept called relative max, there is also the relative min. Well, it’s the
same thing as max, only it’s the smallest point on the graph in a surrounding area. More formally,
x0is a relative min if there is an open interval containint x 0n which f(x ) i0 the smallest value.
Functions can certainly have both relative maxes and mins, perhaps just one but not the other, or
For both, think of f(x) = sin(x). For just a min, think f(x) = x . For just a max, think
f(x) = −x . And for neither maxes nor mins occuring anywhere, think f(x) = x.
Relative maxes and mins are under the giant category of Relative Extrema, by the way. So,
when I talk about relative extrema, I just mean either a relative max or min, whichever.
A critical point is a point in the domain of f at which there is a horizontal tangent (ie, f (x) = 0)
or the derivative doesn’t exist.
A stationary point is a point at which f (x) = 0 only.
In other words, every stationary point is a critical point, but not every critical point is a
Relative extrema only occur at critical points. So, in other words, if you want to ﬁnd the relative
extrema of a function, you have to ﬁnd the critical points! How do you do that? Well, just set the
derivative equal to zero to ﬁnd the stationary points, and then look for where the derivative is
undeﬁned for the rest of ’em!
1 ′ 3
Example: Find the critical points of f(x) where f (x) = x − 4x.
Solution: Well, our derivative is a polynomial, so it’s nice and deﬁned everywhere. This means
we only need to worry about when f (x) = 0 (ie, just the stationary points). Therefore,
′ 3 2
f (x) = x − 4x = x(x − 4) = x(x − 2)(x + 2)
and when we set this equal to zero, x = −2,0,2.
The First Derivative Test:
Ah, so now we can actually ﬁnd out if something is a relative max or min! In fact, we can have
critical points that are not relative extrema! How do we determine this? Why, with the ﬁrst
derivative test of course! So, here’s what the test says: if f changes sign at a critical point, then
there’s a relative extremum there. Moreover, as f (x) goes over a critical point:
- if f (x) goes from positive to negative, then there’s a relative max.
- if f (x) goes from negative to positive, then there’s a relative min.
- if f (x) has the same sign (positive to positive or negative to negative), there is no relative
extremum at that critical point.
I highly highly suggest that you take a look at ﬁgure 5.2.6 on page 281 in your book to get an idea
of diﬀerent types of critical points and which give relative extrema.
The diﬀerent types of critical points that you may run into (at least the ones that give you relative
- horizontal tangents (ok, yeah you know this, but I had to put it in here anyway!)
Gee, it feels like we went over this stuﬀ already when we ﬁrst started studying derivatives! Wait a
minute! We did basically study this stuﬀ already! Remember, the deal with corners and cusps are
that the function is, in fact, deﬁned and continuous at these points, but you get problems with the
derivative. A corner comes from when you get a jump discontinuity in the derivative. A cusp
comes about when you get a VA in the derivative (but everything was ﬁne in the original
2 Example: Find the relative extrema of f(x) where f (x) = x − 4x. 3
Solution: All right, let’s bring this guy back and determine where those maxes and mins are! We
found before that f (x) = x(x + 2)(x − 2), so x = −2,0,2 when f (x) = 0. So, we need to do some
checking of signs! This is going to work just like that concavity stuﬀ we did on monday:
−−−−−−− +++++++ −−−−−−− +++++++
−2 0 2
−3 −1 1 3
So, around x = −2, we’re going from negative to positive. f is decreasing and then increasing.
This corresponds to a relative min! As for x = 0, we’re going from positive to negative, which is a
max. Similarly, x = 2 is also a min.
I’m sorry I can’t do more examples, but there’s a lot of material to get through today. I’ll do more
examples if there is time in class.
The Second Derivative Test:
The name says it all. Take the second derivative of a function f in order to determine relative
extrema. So, if f (x )0= 0 (ie, we have a stationary point x ), the0
- if f (x ) > 0, you have a relative min.
- if f (x0) < 0, you have a relative max.
- if f (x0) = 0, you just wasted your time. x could0be a rel max, or perhaps it could be a relative
min...or maybe neither of them!
Example: Why not bring back that same function f where f (x) = x − 4x! Find relative
extrema using the second derivative test.