Lesson Explainer: Variation Functions Mathematics

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

The variation 𝑉 is a number that measures the amount by which a function 𝑦=𝑓(π‘₯) changes, when π‘₯ changes, within the domain of the function, from π‘Ž to 𝑏, as shown in the diagram.

If we denote the change in 𝑦 by Δ𝑦, we can write the variation as 𝑉=Δ𝑦=𝑓(𝑏)βˆ’π‘“(π‘Ž).

The sign of 𝑉 also tells us whether the function increases (𝑉>0), decreases (𝑉<0), or stays the same (𝑉=0) when π‘₯ changes from π‘Ž to 𝑏. More precisely, if we denote the change in π‘₯ by Ξ”π‘₯=π‘βˆ’π‘Ž, the gradient of the line between the points (π‘Ž,𝑓(π‘Ž)) and (𝑏,𝑓(𝑏)) is π‘š=Δ𝑦Δπ‘₯=𝑓(𝑏)βˆ’π‘“(π‘Ž)π‘βˆ’π‘Ž=π‘‰π‘βˆ’π‘Ž.

Since 𝑏>π‘Ž, the sign of the variation 𝑉 is the same as that of the gradient of the line between these two points.

As an example, consider the constant function 𝑓(π‘₯)=3, and let’s find the variation of this function when π‘₯ changes from 0 to 5: 𝑉=𝑓(5)βˆ’π‘“(0)=3βˆ’3=0.

As we would expect, since the function is a constant and therefore never changes, the variation between π‘₯=0 and π‘₯=5 is zero. This is true of the variation between any two values π‘₯=π‘Ž and π‘₯=𝑏: 𝑉=𝑓(𝑏)βˆ’π‘“(π‘Ž)=3βˆ’3=0.

Now, consider the linear function 𝑓(π‘₯)=3π‘₯+1. Suppose we want to find the variation of this function when π‘₯ changes from 2 to 2.5. Since the slope of this line is π‘š=3, which is positive, we would expect the variation of the function to be positive between these values. We can simply substitute these values to find 𝑉=𝑓(2.5)βˆ’π‘“(2)=(3Γ—2.5+1)βˆ’(3Γ—2+1)=1.5.

This is the variation of the function 𝑓 when π‘₯ changes from 2 to 2.5. Since 𝑉 is positive, we know that the function increases between these two values of π‘₯.

From Ξ”π‘₯=π‘βˆ’π‘Ž, we can express 𝑏 in terms of Ξ”π‘₯ and π‘Ž as 𝑏=π‘Ž+Ξ”π‘₯.

Thus, the variation can be written as 𝑉=𝑓(π‘Ž+Ξ”π‘₯)βˆ’π‘“(π‘Ž).

This is the variation of the function when π‘₯ changes by the amount Ξ”π‘₯ starting from π‘₯=π‘Ž.

For the linear function 𝑓(π‘₯)=3π‘₯+1 when π‘₯ changes from 2 to 2.5 starting from π‘₯=2, it changes by the amount Ξ”π‘₯=2.5βˆ’2=0.5.

Using this, we can also express the variation of the linear function 𝑓 as 𝑉=𝑓(2+0.5)βˆ’π‘“(2)=1.5, which gives us the same result, as expected.

In general, since the amount by which π‘₯ changes, Ξ”π‘₯, is arbitrary, we can use a variable β„Ž to express this with β„Ž=Ξ”π‘₯, and hence the variation 𝑉, as a function of β„Ž. This is known as the variation function of 𝑓(π‘₯).

Definition: The Variation Function

The variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž), where β„Ž represents the change in π‘₯ and 𝑉(β„Ž) represents the variation of the function 𝑓(π‘₯) from π‘₯=π‘Ž to π‘₯=π‘Ž+β„Ž.

In other words, the variation function 𝑉(β„Ž) measures the amount by which the function 𝑓 changes when π‘₯ changes from π‘₯=π‘Ž to π‘₯=π‘Ž+β„Ž, where the variable β„Ž is the amount by which π‘₯ changes.

It is instructive to consider an example in which we identify the variation function algebraically when π‘₯ changes between two arbitrary values.

Example 1: Identifying the Variation Function Algebraically

When π‘₯ changes from π‘₯ to π‘₯, the variation function 𝑉(β„Ž) for 𝑓 is .

Answer

In this example, we want to find 𝑉(β„Ž), the variation function, for an arbitrary function 𝑓(π‘₯) when π‘₯ changes from π‘₯ to π‘₯.

The variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž).

If π‘₯ changes from π‘₯ to π‘₯, we have π‘Ž=π‘₯, the starting position, and β„Ž=π‘₯βˆ’π‘₯, the amount by which π‘₯ changes from π‘₯ to π‘₯. On substituting these into 𝑉(β„Ž), the variation function can be expressed algebraically in terms of π‘₯ and π‘₯ as 𝑉(β„Ž)=𝑓(π‘₯+π‘₯βˆ’π‘₯)βˆ’π‘“(π‘₯)=𝑓(π‘₯)βˆ’π‘“(π‘₯).

Let’s now consider the variation function of a linear function.

Example 2: Finding the Variation Function of a Linear Function

If the function π‘“βˆΆπ‘“(π‘₯)=5π‘₯βˆ’3, then the variation function 𝑉(β„Ž)= at π‘₯=2.

Answer

In this example, we want to determine the variation function of the linear function defined by 𝑓(π‘₯)=5π‘₯βˆ’3 at π‘₯=2. We recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž).

For 𝑓(π‘₯)=5π‘₯βˆ’3 at π‘₯=2, the variation function is therefore 𝑉(β„Ž)=𝑓(2+β„Ž)βˆ’π‘“(2)=5(2+β„Ž)βˆ’3βˆ’(5Γ—2βˆ’3)=5β„Ž+10βˆ’3βˆ’7=5β„Ž.

This tells us that for a linear function, the amount by which the function changes or the variation function 𝑉(β„Ž) is always the same no matter what the starting point is, and it is directly proportional to the amount by which the π‘₯-values change, β„Ž. This is expected as 𝑓 is a linear function, which defines a straight line.

In general, for the linear function 𝑓(π‘₯)=π‘šπ‘₯+𝑐, the variation function at π‘₯=π‘Ž is given by 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž)=π‘š(π‘Ž+β„Ž)+π‘βˆ’(π‘šπ‘Ž+𝑐)=π‘šβ„Ž.

The variation function 𝑉(β„Ž) of a linear function 𝑓(π‘₯)=π‘šπ‘₯+𝑐 is proportional to the gradient π‘š of the line.

Now, let’s look at a few other examples in order to practice and deepen our understanding of variation functions. In the next example, we will find the variation function of a quadratic function.

Example 3: Finding the Variation Function of a Quadratic Function

Determine the variation function 𝑉(β„Ž) for 𝑓(π‘₯)=βˆ’8π‘₯βˆ’5π‘₯βˆ’8 at π‘₯=βˆ’1.

Answer

In this example, we will determine the variation of the quadratic function 𝑓(π‘₯)=βˆ’8π‘₯βˆ’5π‘₯βˆ’8 at π‘₯=βˆ’1.

We recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž).

For 𝑓(π‘₯)=βˆ’8π‘₯βˆ’5π‘₯βˆ’8, the variation function at π‘₯=βˆ’1 is therefore 𝑉(β„Ž)=𝑓(βˆ’1+β„Ž)βˆ’π‘“(βˆ’1)=βˆ’8(β„Žβˆ’1)βˆ’5(β„Žβˆ’1)βˆ’8βˆ’ο€Ίβˆ’8(βˆ’1)βˆ’5Γ—βˆ’1βˆ’8=βˆ’8ο€Ήβ„Žβˆ’2β„Ž+1ο…βˆ’5(β„Žβˆ’1)βˆ’8βˆ’(βˆ’8+5βˆ’8)=βˆ’8β„Ž+11β„Ž.

Now, let’s find the variation of a different quadratic function.

Example 4: Finding the Variation Function of a Quadratic Function

Determine the variation function 𝑉(β„Ž) for 𝑓(π‘₯)=βˆ’4π‘₯βˆ’9π‘₯+9 at π‘₯=βˆ’1.

Answer

In this example, we want to determine the variation of the quadratic function 𝑓(π‘₯)=βˆ’4π‘₯βˆ’9π‘₯+9 at π‘₯=βˆ’1. We recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž).

For 𝑓(π‘₯)=βˆ’4π‘₯βˆ’9π‘₯+9, the variation function at π‘₯=βˆ’1 is therefore 𝑉(β„Ž)=𝑓(βˆ’1+β„Ž)βˆ’π‘“(βˆ’1)=ο€Ίβˆ’4(β„Žβˆ’1)βˆ’9(β„Žβˆ’1)+9ο†βˆ’ο€Ίβˆ’4(βˆ’1)βˆ’9(βˆ’1)+9=βˆ’4β„Ž+8β„Žβˆ’4βˆ’9β„Ž+9+9βˆ’14=βˆ’4β„Žβˆ’β„Ž.

In the next example, we will find the variation function of another quadratic function, but this time we will also evaluate it at a particular value of β„Ž, the change in π‘₯.

Example 5: Finding the Value of the Variation Function of a Quadratic Function

If 𝑉 is the variation function for 𝑓(π‘₯)=π‘₯βˆ’4π‘₯+2, what is 𝑉(βˆ’0.2) when π‘₯=8?

Answer

In this example, we will determine the variation function of the quadratic equation 𝑓(π‘₯)=π‘₯βˆ’4π‘₯+2 at π‘₯=8 and evaluate this function for a change in π‘₯ of β„Ž=βˆ’0.2. We recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž).

The variation function of 𝑓(π‘₯)=π‘₯βˆ’4π‘₯+2 at π‘₯=8 is therefore 𝑉(β„Ž)=𝑓(8+β„Ž)βˆ’π‘“(8)=(8+β„Ž)βˆ’4(8+β„Ž)+2βˆ’ο€Ή8βˆ’4Γ—8+2=β„Ž+16β„Ž+64βˆ’4β„Žβˆ’32+2βˆ’(64βˆ’32+2)=β„Ž+12β„Ž.

We can now evaluate this variation function at β„Ž=βˆ’0.2 to find 𝑉(βˆ’0.2)=(βˆ’0.2)+12(βˆ’0.2)=βˆ’2.36.

This means that for a change in π‘₯ of βˆ’0.2 from π‘₯=8, the given function, 𝑓(π‘₯), decreases by an amount of 2.36 units.

Now, let’s look at an example in which we find the variation function of a trigonometric function.

Example 6: Finding the Variation Function of a Trigonometric Function

Determine the variation function of 𝑓(π‘₯)=π‘₯cos at π‘₯=πœ‹2.

Answer

In this example, we want to find the variation function of a trigonometric function 𝑓(π‘₯)=π‘₯cos at π‘₯=πœ‹2. We recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž).

For 𝑓(π‘₯)=π‘₯cos at π‘₯=πœ‹2, the variation function is therefore 𝑉(β„Ž)=π‘“ο€»πœ‹2+β„Žο‡βˆ’π‘“ο€»πœ‹2=ο€»πœ‹2+β„Žο‡βˆ’ο€»πœ‹2.coscos

Using cosο€»πœ‹2=0 and the cofunction identity, cossinο€»πœ‹2+π‘₯=βˆ’(π‘₯).

The variation function becomes 𝑉(β„Ž)=βˆ’(β„Ž).sin

In the next example, we will find the variation function of an exponential function.

Example 7: Finding the Variation Function of an Exponential Function

Determine the variation function 𝑉(β„Ž) of 𝑓(π‘₯)=π‘’οŠ©ο— at π‘₯=2.

Answer

In this example, we want to determine the variation function of the exponential function 𝑓(π‘₯)=π‘’οŠ©ο— at π‘₯=2. We recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž).

For 𝑓(π‘₯)=π‘’οŠ©ο—, the variation function at π‘₯=2 is therefore 𝑉(β„Ž)=𝑓(2+β„Ž)βˆ’π‘“(2)=π‘’βˆ’π‘’=π‘’βˆ’π‘’=π‘’ο€Ήπ‘’βˆ’1.(οŠ¨οŠ°ο‚)οŠ¬οŠ©ο‚οŠ°οŠ¬οŠ¬οŠ¬οŠ©ο‚

Now, let’s find the variation function of a quadratic function and use this to find the value of an unknown coefficient in the quadratic.

Example 8: Determining the Variation Function of a Quadratic Function and Finding the Value of One of Its Unknowns

Determine the variation function 𝑉(β„Ž) for 𝑓(π‘₯)=βˆ’π‘₯+π‘Žπ‘₯+17 at π‘₯=βˆ’1. Additionally, find π‘Ž if 𝑉49=116.

Answer

In this example, we want to determine the variation function of 𝑓(π‘₯)=βˆ’π‘₯+π‘Žπ‘₯+17 at π‘₯=βˆ’1 then use the given value 𝑉49=116 to determine the unknown coefficient π‘Ž.

Recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=𝑐 is defined as 𝑉(β„Ž)=𝑓(𝑐+β„Ž)βˆ’π‘“(𝑐).

For 𝑓(π‘₯)=βˆ’π‘₯+π‘Žπ‘₯+17, the variation function at π‘₯=βˆ’1 is therefore 𝑉(β„Ž)=𝑓(βˆ’1+β„Ž)βˆ’π‘“(βˆ’1)=βˆ’(βˆ’1+β„Ž)+π‘Ž(βˆ’1+β„Ž)+17βˆ’ο€Ίβˆ’(βˆ’1)βˆ’π‘Ž+17=βˆ’β„Ž+2β„Žβˆ’1+π‘Žβ„Žβˆ’π‘Ž+17βˆ’(βˆ’1βˆ’π‘Ž+17)=βˆ’β„Ž+2β„Ž+π‘Žβ„Ž=βˆ’β„Ž+(2+π‘Ž)β„Ž.

Now, we can determine the coefficient π‘Ž by substituting β„Ž=49 into the variation function: 𝑉49=βˆ’ο€Ό49+49(2+π‘Ž)=βˆ’1681+89+49π‘Ž.

Thus, 𝑉49=116 gives us the equation βˆ’1681+89+49π‘Ž=116.

In order to find the value of π‘Ž, we can multiply this by 94 to give ο€Όβˆ’1681+89+49π‘ŽοˆΓ—94=116Γ—94,βˆ’49+2+π‘Ž=338.

On rearranging this to make π‘Ž the subject, we find π‘Ž=49βˆ’2+338=18572=2.5694….

Therefore, to two decimal places, we have π‘Ž=2.57.

Now, let’s look at an example in which we find the variation function of a trigonometric function and use this to find the value of an unknown coefficient.

Example 9: Finding the Variation Function of a Trigonometric Function and Finding the Value of One of Its Unknowns

Determine the variation function of 𝑓(π‘₯)=π‘Žπ‘₯sin at π‘₯=πœ‹.

If π‘‰ο€»πœ‹2=1, find π‘Ž.

Answer

In this example, we want to find the variation function of 𝑓(π‘₯)=π‘Žπ‘₯sin at π‘₯=πœ‹ then use the given value π‘‰ο€»πœ‹2=1 to determine the unknown coefficient π‘Ž. We recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=𝑐 is defined as 𝑉(β„Ž)=𝑓(𝑐+β„Ž)βˆ’π‘“(𝑐).

For 𝑓(π‘₯)=π‘Žπ‘₯sin, the variation function at π‘₯=πœ‹ can be found to be 𝑉(β„Ž)=𝑓(πœ‹+β„Ž)βˆ’π‘“(πœ‹)=π‘Ž(πœ‹+β„Ž)βˆ’π‘Žπœ‹.sinsin

Using sinπœ‹=0 and the identity sinsin(πœ‹+π‘₯)=βˆ’π‘₯, the variation function becomes 𝑉(β„Ž)=βˆ’π‘Žβ„Ž.sin

If π‘‰ο€»πœ‹2=1, we can substitute this value to find π‘‰ο€»πœ‹2=βˆ’π‘Žο€»πœ‹2=1.sin

Solving for π‘Ž then gives us π‘Ž=βˆ’1.

In the next example, we want to determine an unknown value using a given variation function at that value and comparing it with the variation function that we find directly from the given function.

Example 10: Finding the Values of an Unknown given a Quadratic Function and Its Variation Function

If the variation function of 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯ at π‘₯=𝑑 is 𝑉(β„Ž)=π‘Žβ„Ž+π‘β„ŽοŠ¨, what is the value of 𝑑?

Answer

In this example, we are given the variation function of a particular function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯ at π‘₯=𝑑, and by comparing our result from the formula for a variation of a function with the stated variation function, we will determine the value of 𝑑.

Recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=𝑐 is defined as 𝑉(β„Ž)=𝑓(𝑐+β„Ž)βˆ’π‘“(𝑐).

For 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯, the variation function at π‘₯=𝑑 is therefore 𝑉(β„Ž)=𝑓(𝑑+β„Ž)βˆ’π‘“(𝑑)=π‘Ž(𝑑+β„Ž)+𝑏(𝑑+β„Ž)βˆ’ο€Ήπ‘Žπ‘‘+𝑏𝑑=π‘Žο€Ήβ„Ž+2β„Žπ‘‘+𝑑+𝑏(β„Ž+𝑑)βˆ’ο€Ήπ‘Žπ‘‘+𝑏𝑑=π‘Žβ„Ž+2π‘Žβ„Žπ‘‘+π‘Žπ‘‘+π‘β„Ž+π‘π‘‘βˆ’ο€Ήπ‘Žπ‘‘+𝑏𝑑=π‘Žβ„Ž+π‘β„Ž+2π‘Žβ„Žπ‘‘.

Comparing this result with the given variation function, 𝑉(β„Ž)=π‘Žβ„Ž+π‘β„ŽοŠ¨, we have π‘Žβ„Ž+π‘β„Ž=π‘Žβ„Ž+π‘β„Ž+2π‘Žβ„Žπ‘‘βŸΉ2π‘Žβ„Žπ‘‘=0.

Since π‘Žβ‰ 0 and β„Ž is arbitrary, we must have 𝑑=0.

In the final example, we will find the variation function of a quadratic equation and use this with a given value of the function to determine the value of two unknowns that appear as coefficients of the function.

Example 11: Finding the Variation Function of a Quadratic Function Then Determining the Values of Its Constants

Determine the variation function 𝑉(β„Ž) for 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+2 at π‘₯=1, and, from 𝑉12=72 and 𝑓(1)=6, determine the constants π‘Ž and 𝑏.

Answer

In this example, we want to find the variation function of 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+2 at π‘₯=1 and use 𝑉12=72 and 𝑓(1)=6 to determine the unknown constants π‘Ž and 𝑏.

Recall that the variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=𝑐 is defined as 𝑉(β„Ž)=𝑓(𝑐+β„Ž)βˆ’π‘“(𝑐).

For 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+2, the variation function at π‘₯=1 is therefore 𝑉(β„Ž)=𝑓(1+β„Ž)βˆ’π‘“(1)=π‘Ž(1+β„Ž)+𝑏(1+β„Ž)+2βˆ’(π‘Ž+𝑏+2)=π‘Žο€Ήβ„Ž+2β„Ž+1+𝑏(β„Ž+1)+2βˆ’(π‘Ž+𝑏+2)=π‘Žβ„Ž+2π‘Žβ„Ž+π‘β„Ž=π‘Žβ„Ž+β„Ž(2π‘Ž+𝑏).

We can use 𝑉12=72 and 𝑓(1)=6 to determine the constants π‘Ž and 𝑏 by forming two simultaneous equations. In particular, 𝑓(1)=π‘Ž(1)+𝑏(1)+2=π‘Ž+𝑏+2=6,𝑉12=π‘Žο€Ό12+12(2π‘Ž+𝑏)=5π‘Ž4+𝑏2=72.

Therefore, we have to solve the simultaneous equations π‘Ž+𝑏=4,5π‘Ž+2𝑏=14.

Rearranging the first equation gives us 𝑏=4βˆ’π‘Ž, and substituting this into the second equation gives us 5π‘Ž+2(4βˆ’π‘Ž)=145π‘Ž+8βˆ’2π‘Ž=143π‘Ž+8=14.

Solving this for π‘Ž, we find 3π‘Ž=14βˆ’8=6π‘Ž=2.

Substituting this value back into the first equation gives us 𝑏=4βˆ’π‘Ž=4βˆ’2=2.

Thus, with the variation function 𝑉(β„Ž)=π‘Žβ„Ž+β„Ž(2π‘Ž+𝑏), we find that π‘Ž=2,𝑏=2.

The variation function 𝑉(β„Ž) is also related to the average rate of change 𝐴(β„Ž) of a function 𝑓 defined by 𝐴(β„Ž)=𝑉(β„Ž)β„Ž=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž)β„Ž and the instantaneous rate of change, also known as the first derivative of 𝑓 at π‘₯=π‘Ž, 𝑓′(π‘Ž)=𝑉(β„Ž)β„Ž=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž)β„Ž.limlimο‚β†’οŠ¦ο‚β†’οŠ¦

However, these are beyond the scope of this explainer and will be covered elsewhere in more detail.

Key Points

  • The variation 𝑉 of a function is a number that measures the amount by which a function changes from π‘₯=π‘Ž to π‘₯=𝑏.
  • The sign of the variation indicates in which overall direction a function changes between the two points (π‘Ž,𝑓(π‘Ž)) and (𝑏,𝑓(𝑏)), and it is the same sign as the slope or gradient of the line between these points. In particular, between π‘₯=π‘Ž and π‘₯=𝑏,
    • if 𝑉>0, then 𝑓(π‘₯) is increasing,
    • if 𝑉<0, then 𝑓(π‘₯) is decreasing,
    • if 𝑉=0, then 𝑓(π‘₯) does not change.
  • The variation function 𝑉(β„Ž) of a function 𝑓(π‘₯) at π‘₯=π‘Ž is defined as 𝑉(β„Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž). This is a measure of how much the function changes when π‘₯ changes from π‘Ž to π‘Ž+β„Ž or, in other words, when π‘₯ starts from π‘Ž and changes by a variable amount β„Ž.
  • A variation function can be used to determine an unknown coefficient or starting value π‘₯=π‘Ž for various functions when we are given the variation at a particular value for β„Ž (i.e., 𝑉(β„Ž)=π‘‰οŠ¦οŠ¦), or other information about the function 𝑓.

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