### Video Transcript

A particle moving along a line has
at each time π‘ a position function π of π‘, which is continuous. Assume that π evaluated at two is
equal to negative five and π evaluated at five is equal to negative two. Another particle moves such that
its position is given by β of π‘ is equal to π of π‘ plus π‘. Which of the following must be
true? Option (A) β has at least one
zero. Option (B) π has at least one
zero. Option (C) π of π‘ is greater than
negative five and π of π‘ is less than negative two for all values of π‘ between
two and five. Option (D) There exists at least
one π, where π is greater than two and π is less than five, such that π
evaluated at π is equal to one. Or option (E) there exists at least
one π, where π is greater than two and π is less than five, such that β evaluated
at π is equal to five.

The question tells us the position
of this particle moving along a line at time π‘ is given by the function π of π‘
which is continuous. And weβre also told when π‘ is
equal to two, our function π of π‘ is equal to negative five. And when π‘ is equal to five, our
function π of π‘ is equal to negative two. This is all the information weβre
given about our function π of π‘. Weβre also told thereβs another
particle whose position is given by the function β of π‘, which is equal to π of π‘
plus π‘.

We can use the fact that we know π
evaluated at two and π evaluated at five to find β evaluated at two and β evaluated
at five. We get that β evaluated at two is
negative three and β evaluated at five is three. Also, since we know that our
function π of π‘ is continuous and π‘ is continuous, β of π‘ which is π of π‘ plus
π‘ is the sum of two continuous functions. So, β of π‘ must also be
continuous.

So far, weβve shown that both π of
π‘ and β of π‘ are continuous functions. We were given two values for π of
π‘. And weβll use this to find two
values for β of π‘. Letβs plot these four points onto a
graph. Weβre now ready to start going
through our options. Option (A) tells us that β must
have at least one zero. The first thing we notice about our
two points in the sketch of β of π‘ is one lies below the π₯-axis and the other lies
above the π₯-axis. And remember, we also know our
function β of π‘ is continuous. So, this should remind us of the
intermediate value theorem.

We recall the intermediate value
theorem tells us if a function π is continuous on the closed interval from π to π
and we have some value π which is between π evaluated at π and π evaluated at
π. Then there exists a π in the open
interval from π to π such that π evaluated at π is equal to π. We want to apply the intermediate
value theorem to our function β of π‘ to show that it has at least one zero.

So, weβll set our value of π equal
to zero, our function π to be β. And weβll set the endpoints of our
intervals to be the two points that we know for our function β of π‘. Thatβs two and five. Now, our new version of the
intermediate value theorem tells us if β is continuous on the closed interval from
two to five and zero is between β evaluated at two and β evaluated at five. Then there exists a π on the open
interval from two to five such that β evaluated at π is equal to zero.

In other words, if we can show that
weβre allowed to use the intermediate value theorem for β of π‘, then we can prove
that there exists at least one zero. There are two conditions for using
this version of the intermediate value theorem. First, we need to show that β is
continuous on the closed interval from two to five. And we already showed that this is
true. β of π‘ is the sum of two
continuous functions, so itβs continuous.

The second condition we need to
show is true is zero is between β evaluated at two and β evaluated at five. And of course, we already found
these values; theyβre negative three and three, respectively. And zero is between negative three
and three. So, this condition is also
true. So, both of the conditions for this
version of the intermediate value theorem are true. This means there must exist a π on
the open interval from two to five such that β evaluated at π is equal to zero.

Therefore, option (A) must be
true. β must have at least one zero. We could stop there, but for due
diligence, letβs check all of our other options. Option (B) tells us that our
function π must have at least one zero. To show that this is not true, we
just need to find the continuous function which passes through both of these points
and has no zeros. One such example could be the
following piecewise linear function. Weβll define this function as
negative two if π‘ is greater than or equal to five, π‘ minus seven if π‘ is between
two and five, and negative five if π‘ is less than or equal to two.

We can see from this definition of
π of π‘ that π of two is negative five and π of five is negative two. Itβs also a piecewise linear
function where all of the endpoints meet up. So, π of π‘ is continuous. And we can see that π of π‘ has no
zeros. Therefore, option (B) does not have
to be true. So, now that weβve shown that
option (B) is not true, letβs clear some space and work on option (C).

Option (C) tells us we should be
able to bound our function π of π‘ between negative five and negative two for all
values of π‘ between two and five. We can show that this is also false
by using a similar method we did before. Weβll use a piecewise linear
function. One such example of this is π of
π‘ is equal to negative π‘ over two minus four if π‘ is between zero and four. And π of π‘ is equal to four π‘
minus 22 if π‘ is greater than four.

Again, we can see that π evaluated
at two is equal to negative five and π evaluated at five is equal to negative
two. However, we can also see that π
evaluated at four is equal to negative six. And π of π‘ is a piecewise linear
function whose endpoints meet. So, itβs a continuous function. So, weβve again found a continuous
version of π of π‘ which passes through our two points. However, in our case, when π‘ is
equal to four, π of π‘ is negative six. So, itβs not between negative five
and negative two when π‘ is between two and five.

We donβt actually need to clear our
working for option (D). We can see that option (D) tells us
that there should exist at least one π, where π is between two and five, such that
π evaluated at π is equal to one. However, if we look at our version
of π of π‘, where π‘ is between two and five, we can see the highest output of π
of π‘ is negative two. So, in this case, thereβs no such
π such that π of π is equal to one. So, option (D) is also not
true. Letβs clear our working and look at
the final option, option (E).

Option (E) tells us that there
should exist at least one π between the values of two and five such that β
evaluated at π is equal to five. One such way to show that this is
not true is to take the straight line which passes between our two points, where π‘
is greater than or equal to zero. Thatβs β of π‘ is equal to two π‘
minus seven. And remember that β of π‘ is equal
to π of π‘ plus π‘. So, this also tells us that π of
π‘ is equal to π‘ minus seven. And of course, this version of π
of π‘ is linear. So, itβs continuous and it passes
through our two points.

Letβs look at the largest output of
β of π‘, where π‘ is between two and five; itβs equal to three. If the highest output of our
function is three in this interval, it can never be equal to five.

Therefore, weβve shown for a
continuous function π of π‘, where π of two is equal to negative five and π of
five is equal to negative two and the second function β of π‘ which is equal to π
of π‘ plus π‘. Then, the only one of our options
which must be true is that β must have at least one zero.