# Introduction to Functions

I'm studying Pre-Calculus DeMYSTiFieD by Rhonda Huettenmueller.
I was doing good till Page 31.

My Question is how did they get -1 in there answer?
I don't Understand it looks so simple.

f(x)=1/x

### Comments for Introduction to Functions

 Aug 30, 2011 Newton’s Quotient by: Staff --------------------------------------------------- Part II Computing a common denominator We can compute a common denominator by multiplying the denominators of both fractions: Common denominator = (a+h)*a = a(a+h) The next step is to convert each of the fractions (1/(a+h), and 1/a) to fractions with the denominator of a(a+h). Convert the fraction 1/(a+h) to a fraction with a denominator of a(a+h) The denominator of the fraction 1/(a+h) is (a+h). To convert the denominator of (a+h) to a denominator of a(a+h), we are going to multiply the (a+h) by a. However, simply multiplying the denominator by “a” would change the value of the fraction 1/(a+h) to 1/a(a+h). Since we cannot change the value of the fraction, we are going to multiply both the numerator AND the denominator by a. In other words, the fraction 1/(a+h) will be multiplied by another fraction: a/a. Note that a/a = 1, so we are actually multiplying by 1. The value of the original fraction 1/(a+h) will not change. = [1/(a+h)]*(a/a) When multiplying two fractions, multiply the two numerators and multiply the two denominators: = (1*a)/[(a+h)*a] 1/(a+h) = a/[a(a+h)] Convert the fraction 1/a to a fraction with a denominator of a(a+h) The denominator of the fraction 1/a is a. To convert the denominator of a to a denominator of a(a+h), we are going to multiply the a by (a+h). As before, we cannot change the value of the fraction. We are going to multiply both the numerator AND the denominator by (a+h). In other words, the fraction 1/a will be multiplied by another fraction: (a+h)/(a+h). Note that (a+h)/(a+h) = 1, so we are actually multiplying by 1. The value of the original fraction 1/a will not change. = (1/a)*[(a+h)/(a+h)] When multiplying two fractions, multiply the two numerators and multiply the two denominators: = [1*(a+h)]/ [a*(a+h)] 1/a = (a+h)/[a(a+h)] The slope formula is now: = [1/(a+h) - 1/a]/h = {a/[a(a+h)] - (a+h)/[a(a+h)]}/h Since each fraction has the same denominator, the fractions can finally be subtracted. When subtracting fractions, only the numerators are subtracted. = {[a - (a+h)]/[a(a+h)]}/h = {[a – a - h)]/[a(a+h)]}/h = {[0 - h)]/[a(a+h)]}/h = {[-h)]/[a(a+h)]}/h = {[-h]/[a(a+h)]}/h {[-h]/[a(a+h)]} is actually being divided by the fraction h/1. Therefore, it can be multiplied by 1/h. = {[-h]/[a(a+h)]}*{1/h} = {[-h*1]/ [a(a+h) *h]} = (-h/h)*{1/[a(a+h)]} This is the place the -1 comes from: –h/h = -1 = (-1)*{1/[a(a+h)]} = -1/[a(a+h)] The final answer is: -1/[a(a+h)] --------------------------------------------------- Although you didn’t ask, the final step of the process is: Lim h→0 [f(a + h)- f(a)]/h = Lim h→0 -1/[a(a+h)] = Lim h→0 -1/[a(a+0)] = -1/[a(a)] = -1/a² Since x = a = -1/x² The derivative (slope) of the function f(x)=1/x is: f’(x) = -1/x² Thanks for writing. Staff www.solving-math-problems.com

 Aug 30, 2011 Newton’s Quotient by: Staff Part I The question: I'm studying Pre-Calculus DeMYSTiFieD by Rhonda Huettenmueller. I was doing good till Page 31. My Question is how did they get -1 in there answer? I don't Understand it looks so simple. f(x)=1/x The answer: All of the problems on page 31 demonstrate the preliminary steps to determining the slope of a curve by showing you how to calculate Newton’s Quotient. But before proceeding, let’s back up and review something you are already familiar with: how to calculate the slope of a straight line. The technique used to calculate the slope of a straight line and the technique used to calculate Newton’s Quotient are exactly the same. Suppose you start with the following linear function: f(x) = 2x + 1 You already know the slope of this function is a constant. The function [f(x) = 2x + 1] has only one slope, regardless of the value of x. slope = “rise” over “run”, or “rise” DIVIDED BY “run” = rise/run = [f(x₂) - f(x₁)]/[x₂ - x₁] You can create a table of values and calculate the slope: Since f(x) = 2x + 1 If x₁ = 2, f(x₁) = 2*2+1 = 5 If x₂ = 4, f(x₂) = 2*4+1 = 9 The x-y coordinates of the two points are: (2,5) and (4,9) Slope = [f(x₂) - f(x₁)]/[x₂ - x₁] Slope = [9 - 5]/[4 - 2] Slope = 4/2 Slope = 2 You can now apply the same technique to calculate Newton’s Quotient. The only difference is: Newton’s Quotient only approximates the slope of a curve until the final limit is computed. When the value of x changes, the slope changes. Instead of a linear function such as f(x) = 2x + 1, you have a function like this: f(x) = 1/x. Start with same formula: Slope = rise/run = [f(x₂) - f(x₁)]/[x₂ - x₁] we are going to change the nomenclature, but the formula is not going to change at all x₂ - x₁ = Δx f(x₁) = f(x) f(x₂) = f(x+Δx) the formula for the slope now looks like this: = [f(x+Δx)- f(x)]/Δx To be consistent with the notation in your textbook, we are going to change the notation again, but the formula is still not going to change: Δx = h x = a the formula for the slope now looks like this: = [f(a + h)- f(a)]/h ----------------------------------- YOUR TEXTBOOK STOPS HERE. However, the next step is to take the limit of the slope formula as h approaches 0: Lim h→0 [f(a + h)- f(a)]/h ----------------------------------- Back to your original question: f(x)=1/x How did they get -1 in there answer? f(x)=1/x substitute a+h for the x f(a+h)=1/(a+h) substitute a for the x f(a)=1/a substitute the values for f(a+h) and f(a) into the slope formula = [f(a + h)- f(a)]/h = [1/(a+h) - 1/a]/h Now it’s just a matter of algebra The two fractions 1/(a+h) and 1/a cannot be subtracted unless both fractions have the same denominator. At this point, we have two different denominators: (a+h), and a ---------------------------------------------------