Lesson Video: Variation Functions Mathematics

In this video, we will learn how to evaluate the variation function at a point for a given function.

15:45

Video Transcript

In this video, we’ll learn how to evaluate the variation function at a point for a given function. We’ll 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 𝑦 is equal to 𝑓 of π‘₯. As the value of π‘₯ changes, so does the value of our function 𝑓. And our function is defined on the graph by the points π‘₯, 𝑓 of π‘₯. When π‘₯ is equal to π‘Ž, 𝑦 is equal to 𝑓 of π‘Ž, and we have the point on our graph π‘Ž, 𝑓 of π‘Ž. And when π‘₯ is equal to 𝑏, 𝑦 is 𝑓 of 𝑏, and we have the point on our graph 𝑏, 𝑓 of 𝑏. And the variation 𝑉 measures the change in 𝑓 of π‘₯ when π‘₯ changes from π‘Ž to 𝑏 so that the variation 𝑉 is 𝑓 of 𝑏 minus 𝑓 of π‘Ž. And this is often called Δ𝑦. This corresponds to the change in π‘₯, which is 𝑏 minus π‘Ž, which we call Ξ”π‘₯. If our measure 𝑉 is positive, we consider that 𝑓 of π‘₯ has increased with the change in π‘₯. If 𝑉 is equal to zero, there’s been no change. And if 𝑉 is less than zero, then 𝑓 has decreased with a change in π‘₯.

Suppose, for example, we have the function 𝑦 is equal to 𝑓 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 𝑓 of 𝑏 minus 𝑓 of π‘Ž when π‘₯ changes from π‘Ž to 𝑏. In our case, π‘Ž is equal to two and 𝑏 is equal to four. And our function 𝑓 is four π‘₯ squared plus one. The variation 𝑉 is then 𝑓 of four minus 𝑓 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 𝑓 of π‘₯ changes by an amount of 48 units in the positive direction because 𝑉 is positive.

This means that 𝑓 of π‘₯ has increased by an amount of 48 units. In fact, the sign of the variation 𝑉 is the same as the sign of the slope or gradient of the line between the two points π‘Ž, 𝑓 of π‘Ž and 𝑏, 𝑓 of 𝑏. Now, the variation itself is a number or measure of the change in 𝑓 as π‘₯ changes from π‘Ž to 𝑏. We’re 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 Ξ”π‘₯ the change in π‘₯ from π‘Ž to 𝑏, that’s 𝑏 minus π‘Ž, then we can rearrange this to make 𝑏 the subject so that 𝑏 is π‘Ž plus Ξ”π‘₯. Our variation then becomes 𝑓 of π‘Ž plus Ξ”π‘₯ minus 𝑓 of π‘Ž.

Now Ξ”π‘₯ actually represents an arbitrary change in π‘₯, so we’re going to call this the variable β„Ž. You’ll see this used also in defining derivatives in calculus. We then have the variation function 𝑉 of β„Ž is 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž, which is the change in 𝑓 of π‘₯ when π‘₯ changes by an amount of β„Ž from π‘₯ is equal to π‘Ž. Let’s now see how this works for a linear function.

If the function 𝑓 is such that 𝑓 of π‘₯ is equal to five π‘₯ minus three, then the variation function 𝑉 of β„Ž is equal to what at π‘₯ is equal to two.

We want to find the variation function 𝑉 of β„Ž for 𝑓 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 𝑓 of π‘₯ at π‘₯ is equal to π‘Ž is defined as 𝑉 of β„Ž is 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž, where β„Ž is the change in π‘₯. In our case, our initial point π‘₯ is equal to two, and that’s π‘Ž, so that our variation function 𝑉 of β„Ž is 𝑓 of two plus β„Ž minus 𝑓 of two. Now, substituting first π‘₯ is equal to two plus β„Ž into our function 𝑓, we have 𝑓 of two plus β„Ž is five times two plus β„Ž minus three, that is, 10 plus five β„Ž minus three, which is seven plus five β„Ž.

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

It’s interesting to note that, actually, for any linear function of the form 𝑓 of π‘₯ is π‘šπ‘₯ plus 𝑐, the variation function 𝑉 of β„Ž at π‘₯ is equal to π‘Ž is given by π‘š times π‘Ž plus β„Ž plus 𝑐 minus π‘šπ‘Ž plus 𝑐. And this evaluates to π‘š times β„Ž. That is, 𝑉 of β„Ž is proportional to the slope or gradient π‘š of the line defined by the function 𝑓 of π‘₯. We can also use the variation function to determine the amount by which a function changes for a specific change in π‘₯. Let’s see how this works in an example where, this time, our function 𝑓 of π‘₯ is a quadratic.

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

We’re given a quadratic function 𝑓 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 𝑉 of β„Ž. We can do this using the definition for a function 𝑓 of π‘₯ where its variation function at π‘₯ is equal to π‘Ž is given by 𝑉 of β„Ž is 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž. And that’s where β„Ž is the change in π‘₯. In our case, our function 𝑓 of π‘₯ is π‘₯ squared minus four π‘₯ plus two. Our given value of π‘₯ is eight so that π‘Ž is equal to eight so that our variation function 𝑉 of β„Ž is 𝑓 of eight plus β„Ž minus 𝑓 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„Ž is negative 0.2. And we could substitute this directly into our equation for 𝑉 of β„Ž. Alternatively, we could first find 𝑉 of β„Ž, the variation function in terms of β„Ž, and then substitute β„Ž is negative 0.2. We’re going to use the second method to find the variation function 𝑉 of β„Ž. And to do this, we’re first going to substitute π‘₯ is equal to eight plus β„Ž into our function 𝑓 of π‘₯. And this gives us eight plus β„Ž squared minus four times π‘Ž plus β„Ž plus two. Distributing our parentheses, this gives us 64 plus 16β„Ž plus β„Ž squared minus 32 minus four β„Ž plus two. And collecting like terms, this gives us eight squared plus 12β„Ž plus 34.

Now, to find 𝑓 of eight, we substitute π‘₯ equal to eight into our function 𝑓 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 𝑉 of β„Ž, we have eight squared plus 12β„Ž plus 34 minus 34, that is, β„Ž squared plus 12β„Ž. And this is our variation function 𝑉 of β„Ž for 𝑓 of π‘₯ at π‘₯ is equal to eight. Now, to find 𝑉 of negative 0.2, we substitute negative 0.2 in place of β„Ž. 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. 𝑉 of negative 0.2 for 𝑓 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 𝑓 of π‘₯ decreases by 2.36. It would’ve been slightly quicker to use the direct substitution of β„Ž is negative 0.2 into 𝑉 of β„Ž. However, it is instructive to work through finding the variation function 𝑉 of β„Ž.

Of course, variation functions don’t just apply to polynomial functions. Let’s look now at determining 𝑉 of β„Ž for a trigonometric function. And in this example, we’ll also use the variation function to find an unknown value.

Determine the variation function of 𝑓 of π‘₯ is equal to π‘Ž sin π‘₯ at π‘₯ is equal to πœ‹. If 𝑉 of πœ‹ by two is equal to one, find π‘Ž.

There are two parts to this question. We’re asked to find the variation function for a trigonometric function 𝑓 of π‘₯ is π‘Ž sin π‘₯ at π‘₯ is equal to πœ‹ and then to find the value of π‘Ž using the fact that 𝑉 of πœ‹ by two is equal to one. Let’s begin by recalling that for a function 𝑓 of π‘₯, the variation function 𝑉 of β„Ž at π‘₯ is equal to 𝛼 is 𝑉 of β„Ž is 𝑓 of 𝛼 plus β„Ž minus 𝑓 of 𝛼. And that’s where β„Ž is the change in π‘₯. In our case, we’re given that π‘₯ is equal to πœ‹, and that’s equal to our 𝛼. This means that our variation function 𝑉 of β„Ž is 𝑓 of πœ‹ plus β„Ž minus 𝑓 of πœ‹.

We’re going to need to find then 𝑓 of πœ‹ plus β„Ž and 𝑓 of πœ‹. So first, substituting π‘₯ is equal to πœ‹ plus β„Ž into our function 𝑓, we have 𝑓 of πœ‹ plus β„Ž is equal to π‘Ž times the sin of πœ‹ plus β„Ž. 𝑓 of πœ‹ is simply π‘Ž sin πœ‹ so that our function 𝑉 of β„Ž is π‘Ž sin πœ‹ plus β„Ž minus π‘Ž sin πœ‹. In fact, we know that sin πœ‹ is equal to zero, so π‘Ž sin πœ‹ is zero. And we have 𝑉 of β„Ž is π‘Ž sin πœ‹ plus β„Ž. For our expression sin πœ‹ plus β„Ž, we can use the identity the sin of πœ‹ plus or minus capital 𝐴 is equal to negative or positive the sin of capital 𝐴. In our case, where the angle capital 𝐴 corresponds to β„Ž, we have negative lowercase π‘Ž times the sin of β„Ž. Our variation function 𝑉 of β„Ž is therefore negative π‘Ž sin β„Ž.

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

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

Determine the variation function 𝑉 of β„Ž for 𝑓 of π‘₯ is equal to negative π‘₯ squared plus π‘Žπ‘₯ plus 17 at π‘₯ is equal to negative one. Additionally, find π‘Ž if 𝑉 of four over nine is 11 over six.

We’re given the function 𝑓 of π‘₯ is negative π‘₯ squared plus π‘Žπ‘₯ 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 𝑓 of π‘₯ at π‘₯ is equal to 𝛼, the variation function 𝑉 of β„Ž is 𝑓 of 𝛼 plus β„Ž minus 𝑓 of 𝛼, where β„Ž is the change in π‘₯ from π‘₯ is equal to 𝛼. Once we found our function 𝑉 of β„Ž, we’ll then substitute β„Ž is four over nine to find the value of π‘Ž in our function 𝑓. So we’re given 𝑓 of π‘₯ is negative π‘₯ squared plus π‘Žπ‘₯ plus 17. With π‘₯ is negative one, this is equal to our 𝛼. And substituting 𝛼 is negative one, we have 𝑉 of β„Ž is 𝑓 of negative one plus β„Ž minus 𝑓 of negative one.

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

Our variation function for 𝑓 of π‘₯ is equal to negative π‘₯ squared plus π‘Žπ‘₯ plus 17 at π‘₯ is negative one is equal to 𝑉 of β„Ž is negative β„Ž squared plus β„Ž times π‘Ž plus two. Now we’re given that 𝑉 of four over nine is equal to 11 over six. And this means that if we substitute β„Ž is equal to four over nine into 𝑉 of β„Ž, this should equal 11 over six. And this means that negative four over nine squared plus four over nine times π‘Ž plus two is 11 over six. And we’re going to use this to find the value of π‘Ž. Evaluating this gives us negative 16 over 81 plus four π‘Ž 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‘Ž on the left-hand side.

Canceling through our parentheses, we then have π‘Ž 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 𝑉 of β„Ž is negative β„Ž squared plus β„Ž times π‘Ž plus two, we have a value of π‘Ž equal to 2.57 to two decimal places.

Let’s complete this video by reminding ourselves of some of the key points covered. We know that for a function 𝑓 of π‘₯, the variation 𝑉 measures the change in 𝑓 when π‘₯ changes from π‘₯ is equal to π‘Ž to π‘₯ is equal to 𝑏, and 𝑉 is given by 𝑓 of 𝑏 minus 𝑓 of π‘Ž. We know also that if 𝑉 is positive, that’s greater than zero, 𝑓 of π‘₯ increases between π‘₯ is π‘Ž and π‘₯ is 𝑏. If 𝑉 is zero, there’s no change between π‘₯ is π‘Ž and π‘₯ is 𝑏. And if 𝑉 is less than zero, then 𝑓 of π‘₯ decreases between π‘₯ is π‘Ž and π‘₯ is 𝑏.

The variation function for 𝑓 of π‘₯ at π‘₯ is equal to π‘Ž is 𝑉 of β„Ž is given by 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž, where β„Ž is the change in π‘₯. If 𝑓 of π‘₯ is the linear function π‘šπ‘₯ plus 𝑐, then 𝑉 of β„Ž is π‘š times β„Ž. And this means that for a linear function, the variation function is proportional to the gradient π‘š of the line described by 𝑓 of π‘₯ is π‘šπ‘₯ plus 𝑐. And finally, given a specific change in π‘₯, that’s β„Ž zero, we can use the function evaluated at this point to find information out about the function 𝑓 of π‘₯, for example, unknown coefficients.

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