# 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