### Video Transcript

Which of the following is the graph
of the function π of π₯ equals two π₯ times sin two ππ₯ on the closed interval
negative one to one?

Weβve been given five graphs which
could represent our function two π₯ sin of two ππ₯ over the closed interval
negative one to one. There are a few steps we can take
to decide which of the graphs weβre actually interested in. To begin with, weβre simply going
to set our function equal to zero. This will tell us the location of
any π₯-intercepts. In other words, two π₯ sin of two
ππ₯ is equal to zero.

Well, this function itself is
actually the product of two individual functions. Itβs the product of two π₯ and sin
of two ππ₯. And so for the product of two
functions to be equal to zero, either one or other of those functions must
themselves be equal to zero. So we say either two π₯ is equal to
zero or sin of two ππ₯ is equal to zero.

Letβs divide our first equation by
two so that π₯ is equal to zero. And we can solve our second
equation algebraically, or we can simply consider the shape of the sine curve. We know the graph of π¦ is equal to
sin of π₯ looks a little something like this. We see itβs equal to zero at
negative π, zero, π, and two π.

Now we know that since a sine
function is periodic, these values of π₯ will repeat. Weβll have negative two π,
negative three π, and so on. For now, weβll just consider the
fact that two ππ₯ could be equal to negative two π, negative π, zero, π, and two
π. We divide through by two π. And we find π₯ is equal to negative
one, negative one-half, zero, one-half, and one.

Now, of course, we were only
interested in values of π₯ on the closed interval from negative one to one. So we donβt need to consider any
other points. And weβve found that there are five
π₯-intercepts for our function. They are negative one, negative
one-half, zero, one-half, and one.

Now, in fact, given our graphs,
this would actually be enough to answer the question. But weβll consider some alternative
options. We can consider the value of the
π¦-intercept by setting π₯ equal to zero and finding π of zero. Well, π of zero is zero. And so we now know the location of
the π¦-intercept of our graph. But what about the turning points
or the critical points of our graph?

We recall that the critical points
of a function are points where the first derivative, π prime of π₯, is either equal
to zero or does not exist. So letβs find the first derivative
of our function. Since the function is the product
of two differentiable functions, weβre going to use the product rule. Weβll let π’ be equal to two π₯ and
π£ be equal to sin of two ππ₯. Then dπ’ by dπ₯ is two. We also know that if we
differentiate sin of π₯, we get cos of π₯. So the derivative of π£ with
respect to π₯ is two π times cos of two ππ₯.

The product rule says that the
derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£
by dπ₯ plus π£ times dπ’ by dπ₯. In this case, thatβs two sin of two
ππ₯ plus four ππ₯ times cos of two ππ₯. Now, this is actually defined for
all π₯. So letβs set it equal to zero and
solve for π₯.

This is not a particularly easy
equation to solve. So weβll use the functionality on
our calculator. And when we do, we get π₯ is equal
to positive and negative 0.781 and so on and positive and negative 0.322. Thereβs one other solution, and
thatβs π₯ is equal to zero. So this tells us the π₯-coordinate
of any critical points.

We see that thereβs a total of five
critical points or turning points. And so weβre looking for a graph
which has π₯-intercepts at negative one, negative one-half, zero, one-half, and
one. A π¦-intercept at zero. And five critical points at
positive and negative 0.7819, π₯ equals positive and negative 0.322, and π₯ equals
zero. In fact, the only graph which
satisfies this criteria is (A). So we can say that the graph of the
function π of π₯ is equal to two π₯ sin of two ππ₯ is (A).