Lesson Explainer: Variation Functions | Nagwa Lesson Explainer: Variation Functions | Nagwa

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(𝑥) changes, when 𝑥 changes, within the domain of the function, from 𝑎 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)𝑓(𝑎).

The sign of 𝑉 also tells us whether the function increases (𝑉>0), decreases (𝑉<0), or stays the same (𝑉=0) when 𝑥 changes from 𝑎 to 𝑏. More precisely, if we denote the change in 𝑥 by Δ𝑥=𝑏𝑎, the gradient of the line between the points (𝑎,𝑓(𝑎)) and (𝑏,𝑓(𝑏)) is 𝑚=Δ𝑦Δ𝑥=𝑓(𝑏)𝑓(𝑎)𝑏𝑎=𝑉𝑏𝑎.

Since 𝑏>𝑎, 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(𝑥)=3, and let’s find the variation of this function when 𝑥 changes from 0 to 5: 𝑉=𝑓(5)𝑓(0)=33=0.

As we would expect, since the function is a constant and therefore never changes, the variation between 𝑥=0 and 𝑥=5 is zero. This is true of the variation between any two values 𝑥=𝑎 and 𝑥=𝑏: 𝑉=𝑓(𝑏)𝑓(𝑎)=33=0.

Now, consider the linear function 𝑓(𝑥)=3𝑥+1. Suppose we want to find the variation of this function when 𝑥 changes from 2 to 2.5. Since the slope of this line is 𝑚=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)𝑓(2)=(3×2.5+1)(3×2+1)=1.5.

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

From Δ𝑥=𝑏𝑎, we can express 𝑏 in terms of Δ𝑥 and 𝑎 as 𝑏=𝑎+Δ𝑥.

Thus, the variation can be written as 𝑉=𝑓(𝑎+Δ𝑥)𝑓(𝑎).

This is the variation of the function when 𝑥 changes by the amount Δ𝑥 starting from 𝑥=𝑎.

For the linear function 𝑓(𝑥)=3𝑥+1 when 𝑥 changes from 2 to 2.5 starting from 𝑥=2, it changes by the amount Δ𝑥=2.52=0.5.

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

In general, since the amount by which 𝑥 changes, Δ𝑥, is arbitrary, we can use a variable to express this with =Δ𝑥, and hence the variation 𝑉, as a function of . This is known as the variation function of 𝑓(𝑥).

Definition: The Variation Function

The variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑎 is defined as 𝑉()=𝑓(𝑎+)𝑓(𝑎), where represents the change in 𝑥 and 𝑉() represents the variation of the function 𝑓(𝑥) from 𝑥=𝑎 to 𝑥=𝑎+.

In other words, the variation function 𝑉() measures the amount by which the function 𝑓 changes when 𝑥 changes from 𝑥=𝑎 to 𝑥=𝑎+, where the variable is the amount by which 𝑥 changes.

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

Example 1: Identifying the Variation Function Algebraically

Fill in the blank: When 𝑥 changes from 2 to 2.3, the variation function 𝑉() for 𝑓 becomes .

Answer

In this example, we want to find 𝑉(), the variation function, for an arbitrary function 𝑓(𝑥) when 𝑥 changes from 2 to 2.3.

The variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑎 is defined as 𝑉()=𝑓(𝑎+)𝑓(𝑎).

If 𝑥 changes from 2 to 2.3, we have 𝑎=2, the starting position, and =2.32, the amount by which 𝑥 changes from 2 to 2.3. On substituting these into 𝑉(), the variation function can be expressed algebraically in terms of 2 and 2.3 as 𝑉()=𝑓(2+2.32)𝑓(2)=𝑓(2.3)𝑓(2).

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

Example 2: Finding the Variation Function of a Linear Function

Fill in the blank: If the function 𝑓𝑓(𝑥)=5𝑥3, then the variation function 𝑉()= at 𝑥=2.

Answer

In this example, we want to determine the variation function of the linear function defined by 𝑓(𝑥)=5𝑥3 at 𝑥=2. We recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑎 is defined as 𝑉()=𝑓(𝑎+)𝑓(𝑎).

For 𝑓(𝑥)=5𝑥3 at 𝑥=2, the variation function is therefore 𝑉()=𝑓(2+)𝑓(2)=5(2+)3(5×23)=5+1037=5.

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

In general, for the linear function 𝑓(𝑥)=𝑚𝑥+𝑐, the variation function at 𝑥=𝑎 is given by 𝑉()=𝑓(𝑎+)𝑓(𝑎)=𝑚(𝑎+)+𝑐(𝑚𝑎+𝑐)=𝑚.

The variation function 𝑉() of a linear function 𝑓(𝑥)=𝑚𝑥+𝑐 is proportional to the gradient 𝑚 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() for 𝑓(𝑥)=8𝑥5𝑥8 at 𝑥=1.

Answer

In this example, we will determine the variation of the quadratic function 𝑓(𝑥)=8𝑥5𝑥8 at 𝑥=1.

We recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑎 is defined as 𝑉()=𝑓(𝑎+)𝑓(𝑎).

For 𝑓(𝑥)=8𝑥5𝑥8, the variation function at 𝑥=1 is therefore 𝑉()=𝑓(1+)𝑓(1)=8(1)5(1)88(1)5×18=82+15(1)8(8+58)=8+11.

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() for 𝑓(𝑥)=4𝑥9𝑥+9 at 𝑥=1.

Answer

In this example, we want to determine the variation of the quadratic function 𝑓(𝑥)=4𝑥9𝑥+9 at 𝑥=1. We recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑎 is defined as 𝑉()=𝑓(𝑎+)𝑓(𝑎).

For 𝑓(𝑥)=4𝑥9𝑥+9, the variation function at 𝑥=1 is therefore 𝑉()=𝑓(1+)𝑓(1)=4(1)9(1)+94(1)9(1)+9=4+849+9+914=4.

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 , the change in 𝑥.

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

If 𝑉 is the variation function for 𝑓(𝑥)=𝑥4𝑥+2, what is 𝑉(0.2) when 𝑥=8?

Answer

In this example, we will determine the variation function of the quadratic equation 𝑓(𝑥)=𝑥4𝑥+2 at 𝑥=8 and evaluate this function for a change in 𝑥 of =0.2. We recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑎 is defined as 𝑉()=𝑓(𝑎+)𝑓(𝑎).

The variation function of 𝑓(𝑥)=𝑥4𝑥+2 at 𝑥=8 is therefore 𝑉()=𝑓(8+)𝑓(8)=(8+)4(8+)+284×8+2=+16+64432+2(6432+2)=+12.

We can now evaluate this variation function at =0.2 to find 𝑉(0.2)=(0.2)+12(0.2)=2.36.

This means that for a change in 𝑥 of 0.2 from 𝑥=8, the given function, 𝑓(𝑥), 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(𝑥)=𝑥cos at 𝑥=𝜋2.

Answer

In this example, we want to find the variation function of a trigonometric function 𝑓(𝑥)=𝑥cos at 𝑥=𝜋2. We recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑎 is defined as 𝑉()=𝑓(𝑎+)𝑓(𝑎).

For 𝑓(𝑥)=𝑥cos at 𝑥=𝜋2, the variation function is therefore 𝑉()=𝑓𝜋2+𝑓𝜋2=𝜋2+𝜋2.coscos

Using cos𝜋2=0 and the cofunction identity, cossin𝜋2+𝑥=(𝑥).

The variation function becomes 𝑉()=().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() of 𝑓(𝑥)=𝑒 at 𝑥=2.

Answer

In this example, we want to determine the variation function of the exponential function 𝑓(𝑥)=𝑒 at 𝑥=2. We recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑎 is defined as 𝑉()=𝑓(𝑎+)𝑓(𝑎).

For 𝑓(𝑥)=𝑒, the variation function at 𝑥=2 is therefore 𝑉()=𝑓(2+)𝑓(2)=𝑒𝑒=𝑒𝑒=𝑒𝑒1.()

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() for 𝑓(𝑥)=𝑥+𝑎𝑥+17 at 𝑥=1. Additionally, find 𝑎 if 𝑉49=116.

Answer

In this example, we want to determine the variation function of 𝑓(𝑥)=𝑥+𝑎𝑥+17 at 𝑥=1 then use the given value 𝑉49=116 to determine the unknown coefficient 𝑎.

Recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑐 is defined as 𝑉()=𝑓(𝑐+)𝑓(𝑐).

For 𝑓(𝑥)=𝑥+𝑎𝑥+17, the variation function at 𝑥=1 is therefore 𝑉()=𝑓(1+)𝑓(1)=(1+)+𝑎(1+)+17(1)𝑎+17=+21+𝑎𝑎+17(1𝑎+17)=+2+𝑎=+(2+𝑎).

Now, we can determine the coefficient 𝑎 by substituting =49 into the variation function: 𝑉49=49+49(2+𝑎)=1681+89+49𝑎.

Thus, 𝑉49=116 gives us the equation 1681+89+49𝑎=116.

In order to find the value of 𝑎, we can multiply this by 94 to give 1681+89+49𝑎×94=116×94,49+2+𝑎=338.

On rearranging this to make 𝑎 the subject, we find 𝑎=492+338=18572=2.5694.

Therefore, to two decimal places, we have 𝑎=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(𝑥)=𝑎𝑥sin at 𝑥=𝜋.

If 𝑉𝜋2=1, find 𝑎.

Answer

In this example, we want to find the variation function of 𝑓(𝑥)=𝑎𝑥sin at 𝑥=𝜋 then use the given value 𝑉𝜋2=1 to determine the unknown coefficient 𝑎. We recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑐 is defined as 𝑉()=𝑓(𝑐+)𝑓(𝑐).

For 𝑓(𝑥)=𝑎𝑥sin, the variation function at 𝑥=𝜋 can be found to be 𝑉()=𝑓(𝜋+)𝑓(𝜋)=𝑎(𝜋+)𝑎𝜋.sinsin

Using sin𝜋=0 and the identity sinsin(𝜋+𝑥)=𝑥, the variation function becomes 𝑉()=𝑎.sin

If 𝑉𝜋2=1, we can substitute this value to find 𝑉𝜋2=𝑎𝜋2=1.sin

Solving for 𝑎 then gives us 𝑎=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(𝑥)=𝑎𝑥+𝑏𝑥 at 𝑥=𝑑 is 𝑉()=𝑎+𝑏, what is the value of 𝑑?

Answer

In this example, we are given the variation function of a particular function 𝑓(𝑥)=𝑎𝑥+𝑏𝑥 at 𝑥=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() of a function 𝑓(𝑥) at 𝑥=𝑐 is defined as 𝑉()=𝑓(𝑐+)𝑓(𝑐).

For 𝑓(𝑥)=𝑎𝑥+𝑏𝑥, the variation function at 𝑥=𝑑 is therefore 𝑉()=𝑓(𝑑+)𝑓(𝑑)=𝑎(𝑑+)+𝑏(𝑑+)𝑎𝑑+𝑏𝑑=𝑎+2𝑑+𝑑+𝑏(+𝑑)𝑎𝑑+𝑏𝑑=𝑎+2𝑎𝑑+𝑎𝑑+𝑏+𝑏𝑑𝑎𝑑+𝑏𝑑=𝑎+𝑏+2𝑎𝑑.

Comparing this result with the given variation function, 𝑉()=𝑎+𝑏, we have 𝑎+𝑏=𝑎+𝑏+2𝑎𝑑2𝑎𝑑=0.

Since 𝑎0 and 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() for 𝑓(𝑥)=𝑎𝑥+𝑏𝑥+2 at 𝑥=1, and, from 𝑉12=72 and 𝑓(1)=6, determine the constants 𝑎 and 𝑏.

Answer

In this example, we want to find the variation function of 𝑓(𝑥)=𝑎𝑥+𝑏𝑥+2 at 𝑥=1 and use 𝑉12=72 and 𝑓(1)=6 to determine the unknown constants 𝑎 and 𝑏.

Recall that the variation function 𝑉() of a function 𝑓(𝑥) at 𝑥=𝑐 is defined as 𝑉()=𝑓(𝑐+)𝑓(𝑐).

For 𝑓(𝑥)=𝑎𝑥+𝑏𝑥+2, the variation function at 𝑥=1 is therefore 𝑉()=𝑓(1+)𝑓(1)=𝑎(1+)+𝑏(1+)+2(𝑎+𝑏+2)=𝑎+2+1+𝑏(+1)+2(𝑎+𝑏+2)=𝑎+2𝑎+𝑏=𝑎+(2𝑎+𝑏).

We can use 𝑉12=72 and 𝑓(1)=6 to determine the constants 𝑎 and 𝑏 by forming two simultaneous equations. In particular, 𝑓(1)=𝑎(1)+𝑏(1)+2=𝑎+𝑏+2=6,𝑉12=𝑎12+12(2𝑎+𝑏)=5𝑎4+𝑏2=72.

Therefore, we have to solve the simultaneous equations 𝑎+𝑏=4,5𝑎+2𝑏=14.

Rearranging the first equation gives us 𝑏=4𝑎, and substituting this into the second equation gives us 5𝑎+2(4𝑎)=145𝑎+82𝑎=143𝑎+8=14.

Solving this for 𝑎, we find 3𝑎=148=6𝑎=2.

Substituting this value back into the first equation gives us 𝑏=4𝑎=42=2.

Thus, with the variation function 𝑉()=𝑎+(2𝑎+𝑏), we find that 𝑎=2,𝑏=2.

The variation function 𝑉() is also related to the average rate of change 𝐴() of a function 𝑓 defined by 𝐴()=𝑉()=𝑓(𝑎+)𝑓(𝑎) and the instantaneous rate of change, also known as the first derivative of 𝑓 at 𝑥=𝑎, 𝑓(𝑎)=𝑉()=𝑓(𝑎+)𝑓(𝑎).limlim

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

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