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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

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Aug 30, 2011
Newton’s Quotient
by: Staff

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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)]


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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

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