### Video Transcript

The graph of the first derivative
π prime of a continuous function π is shown. At what values of π₯ does π have a
local maximum and a local minimum?

Now, by looking at the graph, you
might think that the local maxima are at π₯ equals two and π₯ equals five and the
local minima are at π₯ equals three and π₯ equals seven. But we want to know where the
function π has a local minima and maxima. But the graph shown is of π prime,
the first derivative of π, and not π itself.

We found that the local maxima of
π prime are at π₯ equals two and π₯ equals five and the local minima of π prime
are at π₯ equals three and π₯ equals seven. Okay, so itβs not quite as easy as
just reading off the local maxima and minima from the graph that weβve been
given. But we can use this graph of π
prime to find the local maxima and minima of the function π.

First, we use Fermatβs theorem,
which states that if a function π has a local maximum or minimum at π and π prime
of π exists, then π prime of π is zero. Now, we can see that π prime of π
exists for all values of π on the graph. And so Fermatβs theorem it tells us
that we should be looking for values of π for which π prime of π is zero. This corresponds to where the graph
crosses the π₯-axis. So thatβs at π₯ equals one, π₯
equals six, and π₯ equals eight.

Having found the values of π for
which π prime of π is zero, we can read Fermatβs theorem again applying it to our
function. If π has a local maximum or
minimum at π, then π equals one, π equals six, or π equals eight. Therefore, we can see that there
are only three possible values of π, where π could have a local maximum or
minimum.

Fermatβs theorem does not tell us,
however, for each of these three values of π whether π has a local maximum or a
local minimum there or indeed whether it has neither a local maximum nor a local
minimum. For this, we need the first
derivative test. If π prime changes from a positive
to negative at π, then π has a local maximum at π. If π prime changes from negative
to positive at π, then π has a local minimum at π. And if π prime doesnβt change sign
at π, then π has neither a local maximum nor a local minimum at π.

Letβs apply the first derivative
test to our first critical number, π equals one. We know that at π equals one, π
prime of π is zero. But what is the sign of π prime on
either side of π? Well, looking at the graph, we can
see that π prime changes from negative to positive at π equals one. And looking at the first derivative
test, this tells us that π has a local minimum at π equals one.

Okay, what about our second
critical number, π equals six? Again, we look at the graph. But this time, the sign of π prime
is changing from a positive to negative at π equals six. And so there is a local maximum of
π at six.

And finally, we consider our last
critical number, π equals eight, where we see that π prime changes from negative
to positive. And so just like at one, π has a
local minimum at eight. So hereβs our answer then: π has a
local maximum point at π₯ equals six and local minimum points at π₯ equals one and
π₯ equals eight.

We used Fermatβs theorem to show
that we could only have a local maximum or minimum of π at the critical numbers
one, six, and eight, where the derivative of π was zero. And then, we applied the first
derivative test at each of those critical numbers to decide whether there was a
local maximum, a local minimum, or neither there.

Itβs important that you can not
only apply Fermatβs theorem and the first derivative test, but that you can also
remember them and understand intuitively why they are true. Near a local maximum of a function,
the graph of the function looks like this, with the maximum itself appearing at the
very top of the hill. We can see that the tangent to the
graph at this local maximum is horizontal and so has a slope of zero. And so the derivative of the
function at that point must be zero.

So if the local maximum on the
graph is at the point π π of π, then π prime of π is zero. The same is true for a local
minimum. The tangent to the graph at that
point is horizontal and so has a slope of zero. And so the derivative of the
function at that point is zero.

Looking at the sketches, it looks
reasonable then that Fermatβs theorem is true. If π has a local maximum or
minimum at π, then π prime of π, the derivative of π at π, is zero.

Well, how about our first
derivative test? Letβs look at the sketch of our
local maximum. The tangent to the graph for a
value of π₯ less than π has a positive slope, whereas for a value of π₯ just above
π, it has negative slope. So π prime changes from positive
to negative at π, whereas at a local minimum, the sign of π prime changes from
negative to positive.

So if you ever forget which way
around the sign of π prime has to change for that to be a local maximum or a local
minimum using the first derivative test, then donβt panic, just draw a sketch as we
have done now. The key thing to remember is that
the derivative of a function π at a point π₯ is the slope of the tangent to the
graph of the function π at that point.