Calculating the Slope of a Function at x = -2/3 | Independent Study Problem

[morrowcosom]

Homework Statement

We are calculating the slope of the function f(x) = -1/(2x) at x = -2/3.

For the function f(x) = -1/(2x), we now know:

f(-2/3) = 3/4
Next, evaluate the function at a theoretical point x = -2/3+h near -2/3.
We'll use h to symbolize delta x, a small change in x.
f(-2/3+h) =

Homework Equations

The Attempt at a Solution

= -1/(2(-2/3+h))
= -1/(-4/3+2h)
= 3/4-(1/2h)
I am doing independent study using a computer program and it is telling me that I am incorrect. What is wrong with my work?

 

[Mark44]

Homework Statement

We are calculating the slope of the function f(x) = -1/(2x) at x = -2/3.

For the function f(x) = -1/(2x), we now know:

f(-2/3) = 3/4
Next, evaluate the function at a theoretical point x = -2/3+h near -2/3.
We'll use h to symbolize delta x, a small change in x.
f(-2/3+h) =

Homework Equations

The Attempt at a Solution

= -1/(2(-2/3+h))
= -1/(-4/3+2h)
= 3/4-(1/2h)
I am doing independent study using a computer program and it is telling me that I am incorrect. What is wrong with my work?

The next-to-last line looks fine, but could be rewritten as
= -1/(-4/3+2h)
= 1/(4/3 - 2h)

How did you get 3/4 - 1/(2h)?

 

[Mark44]

Now I see what you did.
\frac{1}{\frac{a}{b} + \frac{c}{d}} \neq \frac{b}{a} + \frac{d}{c}

Your weakness in fractions is preventing you from successfully completing these problems.

 

[morrowcosom]

I am a college student that is halfway through a master's degree in business management. I have not taken a basic algebra course in over six years. I am also dirt broke because I have no job, hence I cannot afford an algebra textbook to help me on my fractions. Where would be a good website to go to to learn the types of fraction simplication used in these problems. Right now I am on hiatus from school do to medical issues, and want to get good at math.

Thanks

 

[Mark44]

I've heard a lot of good things about Khan Academy (http://www.khanacademy.org). The main page is divided up into sections. Take a look at the Pre-Algebra section - I saw a couple of links that discuss fraction multiplication and division.

There might be a lot of topics that would be of use to you. Any time you spend at refreshing your math skills will be a good investment, IMO.

Best of luck to you!
Mark

 

[morrowcosom]

I decided to try out some of the new division stuff I saw on Khan by further simplifying a previous problem:

1/(4/3-2h)
(I went ahead and inverted and multiplied the 2nd terms)

(1/1)(3/4)= 3/4
(1/1)(-1/2h)= -1/2h

to get:
= 3/4-1/2h
How does this look?


I was messing around with latex:
(\frac{1}{1})(\frac{3}{4}) = (\frac{3}{4})

(\frac{1}{1})(\frac{-1}{2h}) = (\frac{1}{-2h})

= (\frac{3}{4}) - (\frac{-1}{2h})

 

[Mark44]

I decided to try out some of the new division stuff I saw on Khan by further simplifying a previous problem:

1/(4/3-2h)
(I went ahead and inverted and multiplied the 2nd terms)

(1/1)(3/4)= 3/4
(1/1)(1/-2h)= -1/2h

to get:
3/4-1/2h
How does this look?

It's not right. 1/(4/3) = 3/4 and 1/(-2h) = -1/(2h), but 1/(4/3 - 2h) is not equal to 3/4 - 1/(2h).
IOW, there is no property of fractions that says that the reciprocal of a sum (or difference) is equal to the sum (difference) of the reciprocals.

For example 1/(2 + 3) = 1/5, but 1/2 + 1/3 = 3/6 + 2/6 = 5/6.

If it doesn't work for ordinary numbers, it's not going to work with algebraic expressions.

Also I was messing around with latex, trying to create 1/1, but in the latex format, and ended up with this.

In the LaTeX below you have \1 in the numerator. It should be just 1, like this:
\frac{1}{1}

\frac{\1}{1}
I typed it directly on this page. I was wondering what the error in my formatting was.

I used this link:
https://www.physicsforums.com/misc/howtolatex.pdf

 

[morrowcosom]

It's not right. 1/(4/3) = 3/4 and 1/(-2h) = -1/(2h), but 1/(4/3 - 2h) is not equal to 3/4 - 1/(2h).
IOW, there is no property of fractions that says that the reciprocal of a sum (or difference) is equal to the sum (difference) of the reciprocals.

For example 1/(2 + 3) = 1/5, but 1/2 + 1/3 = 3/6 + 2/6 = 5/6.

According to the 1/(2+3)=1/5...example, I guess since there is no way to subtract 2h from 4/3, then there is no way to take the reciprocal of the combined terms, then there is no way to further simplify it. I solved a point on a curve slope problem using this form as one of my functions. If there is a way to further simplify the problem I would love to know what it is, but I have no idea on Earth how.

 

[Mark44]

The only thing you can do is 2(2/3 - h).

 

[morrowcosom]

All this because I tried to overcomplicate something.

 

[Mark44]

So to finish this off, you have f(x) = -1/(2x), and you want to find the derivative at x = -2/3. I.e., you want f'(-2/3).

f'(-2/3) = \lim_{h \to 0} \frac{f(-2/3 + h) - f(-2/3}{h}
= \lim_{h \to 0} (1/h)(\frac{1}{4/3 - 2h} - 3/4)
= \lim_{h \to 0} (1/h)(\frac{4 - 3(4/3 - 2h)}{4(4/3 - 2h)} )
= \lim_{h \to 0} (1/h)(\frac{6h}{16/3 - 8h})
= \lim_{h \to 0} \frac{6}{16/3 - 2h} = \frac{6}{16/3} = 6 \cdot \frac{3}{16}= 18/16 = 9/8