Functions [f(x)] and substituton

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Homework Help Overview

The discussion revolves around evaluating and simplifying the expression \((f(x+h) - f(x))/h\) for the function \(f(x) = x^2 - 2x\). Participants are exploring the implications of substituting \(x+h\) into the function and the resulting expressions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Some participants attempt to substitute \(f(x)\) directly into the expression, while others question the correctness of their substitutions. There are discussions about the proper form of \(f(x+h)\) and the simplifications that follow.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts and clarifying the correct substitutions. Some guidance has been offered regarding the need to replace \(x\) with \(x+h\) in the function, and there is a recognition of the importance of this concept in future calculus studies.

Contextual Notes

Participants express uncertainty about their substitutions and simplifications, indicating a need for further clarification on the function's behavior when modified by \(h\). There is also mention of the relevance of this concept in upcoming calculus topics.

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Homework Statement


Evaluate and simpliy [f(x+h) - f(x)]/h if f(x) = x^2 -2x


Homework Equations


None


The Attempt at a Solution


I just plugged in f(x), or x^2 - 2x for f(X+h) and f(x) and burned it down to h/h, or 1.
That sounds too easy, though...want to know if it's correct.
 
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You should show your work. If f(x)=x^2-2x, then what is f(x+h)?
 
if f(x) = 2x + 1
then f(x+h) = 2(x+h) + 1
 
cristo said:
You should show your work. If f(x)=x^2-2x, then what is f(x+h)?

Would it simply be x^2 - 2x + h?
(That may be where the problem is...I'm not too sure on that)

If so, I just did:

x^2 - 2x + h - x^2 + 2x --> h ---numerator
h --- denominator

h/h = 1?

EDIT - I see what you did...(rocophysics)
Thanks! I'll try it out.
 
Last edited:
No, it wouldn't. The variable in f(x) is x, whereas the variable in f(x+h) is (x+h): i.e. you need to write (x+h) in place of x in the original function.
 
If f(x) = x^2+2x then what would say f(3) be? f(3)=(3)^2+2(3) right? So what is f(x+h)?
 
If f(x) = x^2, then f(x+h) = (x+h)^2 = x^2 + 2xh + h^2
 
Feldoh said:
If f(x) = x^2+2x then what would say f(3) be? f(3)=(3)^2+2(3) right? So what is f(x+h)?

Right...so f(x+h) is:

(x+h)^2 - 2x + 2h
I foiled out everything, and I think I'm good to go :)
Thanks for the help!
 
Tekee said:
Right...so f(x+h) is:

(x+h)^2 - 2x + 2h
I foiled out everything, and I think I'm good to go :)
Thanks for the help!

f(x) = 2(x)
f(3) = 2(3)
f(x+h) = 2(?)

Other then that (x+h)^2 looks right. :)

All f(x+h) means is go to the function and every where you see an "x" replace it with "x+h"
 
  • #10
better learn this good! you'll encounter it around the ~2nd chapter of your Calculus book.
 
  • #11
rocophysics said:
better learn this good! you'll encounter it around the ~2nd chapter of your Calculus book.

^^He's got a point [f(x+h) - f(x)]/h will be revisited -- A LOT
 
  • #12
Tekee said:
Right...so f(x+h) is:

(x+h)^2 - 2x + 2h
I foiled out everything, and I think I'm good to go :)
Thanks for the help!

But you're original function was x^2-2x so replacing x with x+h you should have
(x+h)^2-2(x+h). Can you expand this? You are off by a sign somewhere in your post, do you see where?
 
  • #13
i remember our first test ... lots of ppl did bad b/c of this concept

they sure were happy when they learned how to take the derivative :)
 

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