- What are some algebra techniques that allow us to simplify the equation of a secant line?
- Why is this important?
Introduction
Given the graph of a function , we have discussed methods to determine the slope of the secant line between two points, and , on the graph. We know that this slope represents the average rate of change of the function on the interval , denoted by . Both of these can be rewritten by letting , so that we have the value representing the horizontal distance between the points. This means that as , the secant line, or the average rate of change of the function, approaches a value known as the slope of the tangent line of at . This will be discussed extensively in future calculus courses, but in this section we will focus on tools to simplify the expression , as they are essential to calculating this limit.
Definitions and examples
Recall the special formula for difference of squares, . For non-square values of and we can use the same idea to rationalize differences (or sums) of square roots through multiplication by the corresponding sum (or difference), which we call the conjugate. Given any expression , real numbers, the conjugate of this expression is . Multiplying such an expression by its conjugate rationalizes it through the distributive property:
The sum is the conjugate of the difference . Likewise, the difference is the conjugate of the sum .
Multiplying such an expression by its conjugate will rationalize the expression.
- (a)
- ,
We consider to avoid any potential undefined values plugged into our function since its domain is . Observe that we cannot combine any terms in (sqrtDQ), but the numerator is of the form . Hence, we will multiply by the conjugate to rationalize the numerator: Remember, in (multbyC), that in order to avoid changing the value of the expression, we must multiply by the conjugate over itself, i.e., multiply by 1. Then (diffSq) has a difference of squares in the numerator and is equal to by cancelling out the in the numerator and the denominator.
This expression that we have found is now in a form that allows us to consider what happens when by removing the from the denominator.
- (b)
- , .We consider to avoid any potential undefined values plugged into our function since its domain is . Once again, we cannot combine any terms in the numerator of (sqDQ), so we will multiply by the conjugate to rationalize the numerator, hoping we will be able to simplify the equation. (sqDQ) is equal to Now, we simply cancel the in the numerator and the denominator, giving
- (c)
- Note that is defined for all real numbers , so we need not worry about the values of and plugged into the difference quotient formula. Now, we can expand out using the angle sum formula for cosine:
Here we can plug in decreasing values of for cosine and sine and start to notice a pattern.
Explore this further by changing the and values in the following Desmos graph: - (d)
- We will consider two regions and ranges of : (1) with and (2) with .
Let’s start with region (1), where and . From this, we know that , so . Hence we have
Alternatively, if we consider region (2), where and , then we have , so that
Notice that this is not as clear-cut if we consider say and . Then we would need to consider if is large enough that . Let’s explore this some more.
Let and . Further, assume , then
Alternatively, if , then As when and . - (e)
- First note that the denominator of our function is greater than zero for all real values of , so the function is defined for all real numbers. Thus, we may calculate the difference quotient without concern for input values of and . This expression for the difference quotient looks rather messy, so let’s find the common denominator and see if we can cancel out some terms in the numerator by combining the fractions. We will leave the terms in the denominator in their current format, but multiply out the in the numerator for ease of simplification.
Note that the common denominator is . Then we have
Observe that we have combined the denominators now and have many common terms in the numerator that can be subtracted from each other, so that Now, all the terms in the numerator have a factor of , so we can cancel the in the numerator and denominator for a final, simplified difference quotient of