### Video Transcript

In this video, weβll learn how to
evaluate the variation function at a point for a given function. Weβll see first how to define a
variation function for a given function and then how to evaluate the variation
function at a point. You should already be familiar with
various types of functions, including polynomials and trigonometric functions.

Suppose we have a function π¦ is
equal to π of π₯. As the value of π₯ changes, so does
the value of our function π. And our function is defined on the
graph by the points π₯, π of π₯. When π₯ is equal to π, π¦ is equal
to π of π, and we have the point on our graph π, π of π. And when π₯ is equal to π, π¦ is
π of π, and we have the point on our graph π, π of π. And the variation π measures the
change in π of π₯ when π₯ changes from π to π so that the variation π is π of
π minus π of π. And this is often called Ξπ¦. This corresponds to the change in
π₯, which is π minus π, which we call Ξπ₯. If our measure π is positive, we
consider that π of π₯ has increased with the change in π₯. If π is equal to zero, thereβs
been no change. And if π is less than zero, then
π has decreased with a change in π₯.

Suppose, for example, we have the
function π¦ is equal to π of π₯, which is four π₯ squared plus one. We want to find the variation when
π₯ changes from π₯ is equal to two to π₯ is equal to four. Now, remember that the variation of
a function is π of π minus π of π when π₯ changes from π to π. In our case, π is equal to two and
π is equal to four. And our function π is four π₯
squared plus one. The variation π is then π of four
minus π of two. And substituting four and two into
our function, we have four times four squared plus one minus four times two squared
plus one. This evaluates to 65 minus 17,
which is 48. So if π₯ changes from two to four,
our function π of π₯ changes by an amount of 48 units in the positive direction
because π is positive.

This means that π of π₯ has
increased by an amount of 48 units. In fact, the sign of the variation
π is the same as the sign of the slope or gradient of the line between the two
points π, π of π and π, π of π. Now, the variation itself is a
number or measure of the change in π as π₯ changes from π to π. Weβre now going to see how the
variation function is formed. And to do this, we can look at the
change in π₯ in a slightly different way. If we call Ξπ₯ the change in π₯
from π to π, thatβs π minus π, then we can rearrange this to make π the subject
so that π is π plus Ξπ₯. Our variation then becomes π of π
plus Ξπ₯ minus π of π.

Now Ξπ₯ actually represents an
arbitrary change in π₯, so weβre going to call this the variable β. Youβll see this used also in
defining derivatives in calculus. We then have the variation function
π of β is π of π plus β minus π of π, which is the change in π of π₯ when π₯
changes by an amount of β from π₯ is equal to π. Letβs now see how this works for a
linear function.

If the function π is such that π
of π₯ is equal to five π₯ minus three, then the variation function π of β is equal
to what at π₯ is equal to two.

We want to find the variation
function π of β for π of π₯ is five π₯ minus three at π₯ is equal to two. To do this, we recall that the
variation function for a function π of π₯ at π₯ is equal to π is defined as π of
β is π of π plus β minus π of π, where β is the change in π₯. In our case, our initial point π₯
is equal to two, and thatβs π, so that our variation function π of β is π of two
plus β minus π of two. Now, substituting first π₯ is equal
to two plus β into our function π, we have π of two plus β is five times two plus
β minus three, that is, 10 plus five β minus three, which is seven plus five β.

Next, if we substitute π₯ is equal
to two into our function π of π₯, we have π of two is five times two minus three,
which is 10 minus three, and thatβs seven. With these values in our variation
function, this gives us π of β is seven plus five β minus seven. And since seven minus seven is
zero, thatβs equal to five β. The variation function π of β for
the function π of π₯ is five π₯ minus three at π₯ is equal to two is therefore π
of β is equal to five β.

Itβs interesting to note that,
actually, for any linear function of the form π of π₯ is ππ₯ plus π, the
variation function π of β at π₯ is equal to π is given by π times π plus β plus
π minus ππ plus π. And this evaluates to π times
β. That is, π of β is proportional to
the slope or gradient π of the line defined by the function π of π₯. We can also use the variation
function to determine the amount by which a function changes for a specific change
in π₯. Letβs see how this works in an
example where, this time, our function π of π₯ is a quadratic.

If π is the variation function for
π of π₯ is equal to π₯ squared minus four π₯ plus two, what is π of negative 0.2
when π₯ is equal to eight?

Weβre given a quadratic function π
of π₯ is π₯ squared minus four π₯ plus two and asked to find the value of the
variation function if π₯ changes from π₯ is equal to eight by an amount of negative
0.2. The first thing we need to do then
is to find the variation function π of β. We can do this using the definition
for a function π of π₯ where its variation function at π₯ is equal to π is given
by π of β is π of π plus β minus π of π. And thatβs where β is the change in
π₯. In our case, our function π of π₯
is π₯ squared minus four π₯ plus two. Our given value of π₯ is eight so
that π is equal to eight so that our variation function π of β is π of eight plus
β minus π of eight.

There are two ways we could
approach this problem. We know that our change in π₯ is
negative 0.2, and this means that β is negative 0.2. And we could substitute this
directly into our equation for π of β. Alternatively, we could first find
π of β, the variation function in terms of β, and then substitute β is negative
0.2. Weβre going to use the second
method to find the variation function π of β. And to do this, weβre first going
to substitute π₯ is equal to eight plus β into our function π of π₯. And this gives us eight plus β
squared minus four times π plus β plus two. Distributing our parentheses, this
gives us 64 plus 16β plus β squared minus 32 minus four β plus two. And collecting like terms, this
gives us eight squared plus 12β plus 34.

Now, to find π of eight, we
substitute π₯ equal to eight into our function π of π₯, which gives us eight
squared minus four times eight plus two. And this evaluates to 34. Now, with these two results into
our function π of β, we have eight squared plus 12β plus 34 minus 34, that is, β
squared plus 12β. And this is our variation function
π of β for π of π₯ at π₯ is equal to eight. Now, to find π of negative 0.2, we
substitute negative 0.2 in place of β. This gives us negative 0.2 squared
plus 12 times negative 0.2. That is 0.04 minus 2.4, which is
negative 2.36. π of negative 0.2 for π of π₯ is
equal to π₯ squared minus four π₯ plus two when π₯ is equal to eight is therefore
negative 2.36.

What this means is that if π₯
changes by an amount of negative 0.2 from π₯ is equal to eight, then the value of
the function π of π₯ decreases by 2.36. It wouldβve been slightly quicker
to use the direct substitution of β is negative 0.2 into π of β. However, it is instructive to work
through finding the variation function π of β.

Of course, variation functions
donβt just apply to polynomial functions. Letβs look now at determining π of
β for a trigonometric function. And in this example, weβll also use
the variation function to find an unknown value.

Determine the variation function of
π of π₯ is equal to π sin π₯ at π₯ is equal to π. If π of π by two is equal to one,
find π.

There are two parts to this
question. Weβre asked to find the variation
function for a trigonometric function π of π₯ is π sin π₯ at π₯ is equal to π and
then to find the value of π using the fact that π of π by two is equal to
one. Letβs begin by recalling that for a
function π of π₯, the variation function π of β at π₯ is equal to πΌ is π of β is
π of πΌ plus β minus π of πΌ. And thatβs where β is the change in
π₯. In our case, weβre given that π₯ is
equal to π, and thatβs equal to our πΌ. This means that our variation
function π of β is π of π plus β minus π of π.

Weβre going to need to find then π
of π plus β and π of π. So first, substituting π₯ is equal
to π plus β into our function π, we have π of π plus β is equal to π times the
sin of π plus β. π of π is simply π sin π so
that our function π of β is π sin π plus β minus π sin π. In fact, we know that sin π is
equal to zero, so π sin π is zero. And we have π of β is π sin π
plus β. For our expression sin π plus β,
we can use the identity the sin of π plus or minus capital π΄ is equal to negative
or positive the sin of capital π΄. In our case, where the angle
capital π΄ corresponds to β, we have negative lowercase π times the sin of β. Our variation function π of β is
therefore negative π sin β.

Next, weβre asked to find the value
of π if π of π by two is equal to one. We can do this by substituting β is
equal to π by two into π of β and putting this equal to one. We then have π of π by two is
equal to negative π sin π by two is equal to one. In fact, we know that sin π by two
is equal to one so that we have negative π times one is equal to one. That is, negative π is equal to
one. And multiplying both sides by
negative one, we have π equal to negative one. The variation function of π of π₯
is equal to π sin π₯ at π₯ is equal to π is π of β is negative π sin β. And if π of π by two is equal to
one, then π is equal to negative one.

In our final example, weβre again
going to determine the variation function for a quadratic. And then weβre going to use a
specific value of the variation function to find an unknown.

Determine the variation function π
of β for π of π₯ is equal to negative π₯ squared plus ππ₯ plus 17 at π₯ is equal
to negative one. Additionally, find π if π of four
over nine is 11 over six.

Weβre given the function π of π₯
is negative π₯ squared plus ππ₯ plus 17 and asked to find the variation function
for this at π₯ is negative one. To do this, we recall that for a
function π of π₯ at π₯ is equal to πΌ, the variation function π of β is π of πΌ
plus β minus π of πΌ, where β is the change in π₯ from π₯ is equal to πΌ. Once we found our function π of β,
weβll then substitute β is four over nine to find the value of π in our function
π. So weβre given π of π₯ is negative
π₯ squared plus ππ₯ plus 17. With π₯ is negative one, this is
equal to our πΌ. And substituting πΌ is negative
one, we have π of β is π of negative one plus β minus π of negative one.

And now evaluating our function π
at π₯ is equal to negative one plus β, we have negative negative one plus β squared
plus π times negative one plus β plus 17. That is, π of negative one plus β
is equal to negative one minus two β plus β squared minus π plus πβ plus 17. And multiplying our parentheses by
negative one and collecting like terms, we have negative β squared plus β times π
plus two plus 16 minus π. Now, evaluating π at π₯ is equal
to negative one, we have negative negative one squared plus π times negative one
plus 17, that is, negative one minus π plus 17, which is 16 minus π. And substituting our two results
into our function π of β, we see that 16 minus π minus 16 minus π is equal to
zero so that π of β is equal to negative β squared plus β times π plus two.

Our variation function for π of π₯
is equal to negative π₯ squared plus ππ₯ plus 17 at π₯ is negative one is equal to
π of β is negative β squared plus β times π plus two. Now weβre given that π of four
over nine is equal to 11 over six. And this means that if we
substitute β is equal to four over nine into π of β, this should equal 11 over
six. And this means that negative four
over nine squared plus four over nine times π plus two is 11 over six. And weβre going to use this to find
the value of π. Evaluating this gives us negative
16 over 81 plus four π over nine plus eight over nine is equal to 11 over six. And now if we add 16 over 81 and
subtract eight over nine from both sides and multiplying both sides by nine over
four, we can isolate π on the left-hand side.

Canceling through our parentheses,
we then have π is equal to 33 over eight plus four over nine minus two, which
evaluates to 2.5694 and so on. So with our variation function π
of β is negative β squared plus β times π plus two, we have a value of π equal to
2.57 to two decimal places.

Letβs complete this video by
reminding ourselves of some of the key points covered. We know that for a function π of
π₯, the variation π measures the change in π when π₯ changes from π₯ is equal to
π to π₯ is equal to π, and π is given by π of π minus π of π. We know also that if π is
positive, thatβs greater than zero, π of π₯ increases between π₯ is π and π₯ is
π. If π is zero, thereβs no change
between π₯ is π and π₯ is π. And if π is less than zero, then
π of π₯ decreases between π₯ is π and π₯ is π.

The variation function for π of π₯
at π₯ is equal to π is π of β is given by π of π plus β minus π of π, where β
is the change in π₯. If π of π₯ is the linear function
ππ₯ plus π, then π of β is π times β. And this means that for a linear
function, the variation function is proportional to the gradient π of the line
described by π of π₯ is ππ₯ plus π. And finally, given a specific
change in π₯, thatβs β zero, we can use the function evaluated at this point to find
information out about the function π of π₯, for example, unknown coefficients.