Cuse these difference quotients

  • Thread starter leftyguitarjo
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In summary: The function you are given is a function of x, not a function of y. So in order to evaluate the quotient, you need to substitute the given function into the difference quotient.
  • #1
leftyguitarjo
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Cuse these difference quotients!

Homework Statement


http://www.webassign.net/www14/symImages/5/8/30f1e141e81c6a3333af41547f8ff2.gif

Evaluate the difference quotient for the given function

http://www.webassign.net/www14/symImages/4/6/bda668fa415dbf5d0c9c2ba4c031d2.gif

Homework Equations


n/a

The Attempt at a Solution


I seem to have the wrong mindset while doing these.

I got 0.

Actually, I don't really know exactly what I'm even going for here.
 
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  • #2


It's not zero. It's a function of x. Why don't you show us what you did? Just substitute the given f(x) into the difference quotient.
 
  • #3


My main thing is that I don't know what the outcome should look like.

It was during the last few weeks of my senior year of high school so I didnt exactly make it to class every day and I missed this stuff. Its coming back to bite me in college now.
 
  • #4


I told you, just substitute the form for f(x) into the difference quotient and try and do some algebra to simplify it. What is f(5)? This isn't a big conceptual problem.
 
  • #5


leftyguitarjo said:

Homework Statement


http://www.webassign.net/www14/symImages/5/8/30f1e141e81c6a3333af41547f8ff2.gif

Evaluate the difference quotient for the given function

http://www.webassign.net/www14/symImages/4/6/bda668fa415dbf5d0c9c2ba4c031d2.gif


Homework Equations


n/a


The Attempt at a Solution


I seem to have the wrong mindset while doing these.

I got 0.

Actually, I don't really know exactly what I'm even going for here.

Later on you will see that the slope of the tangent line at a point say P(a,f(a)), is given by:

[tex]\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a)[/tex]

And only the quotient [tex] \frac{f(x)-f(a)}{x-a}[/tex] is the slope of secat line.

So, you are given a function f(x) and then asked to find the quotient.

[tex]\frac{f(x)-f(5)}{x-5}[/tex] so what u want to do is replace f(x) with what you are given, in other words, write the function f(x)=... whatever it equals, and then for f(5), you first want to evalulate f(5)=...whatever it equals. You evaluate f(5) by plugging 5 in

[tex]f(x)=\frac{x+7}{x+5}[/tex]
 
Last edited by a moderator:
  • #6


leftyguitarjo said:
My main thing is that I don't know what the outcome should look like.

It was during the last few weeks of my senior year of high school so I didnt exactly make it to class every day and I missed this stuff. Its coming back to bite me in college now.

Dick said "Why don't you show us what you did? " Why will you not do that?
 

1. What is a "difference quotient"?

A difference quotient is a mathematical expression that represents the slope of a curve at a specific point. It is used to calculate the rate of change of a function.

2. How is a difference quotient calculated?

A difference quotient is calculated by taking the difference between two points on a function and dividing it by the difference in their x-values. This can be represented as (f(x+h) - f(x)) / h, where h is the change in x-values.

3. What is the significance of using difference quotients in scientific research?

Difference quotients are important in scientific research because they allow us to analyze and understand the behavior of a function at a specific point. They can be used to make predictions and draw conclusions about the rate of change of a system or process.

4. Can difference quotients be used for non-linear functions?

Yes, difference quotients can be used for non-linear functions as well. However, the value of the difference quotient may vary at different points on the curve.

5. Are there any real-world applications of difference quotients?

Yes, difference quotients have several real-world applications, including in physics, economics, and engineering. They are used to analyze the rate of change of physical quantities, such as velocity and acceleration, as well as in financial modeling and optimization problems.

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