Help for: f(x)=3x^2-3 f(x+deltax)-f(x) all over deltax
- Thread starter robbob
- Start date
- Joined
- Apr 12, 2005
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- 11,339
You are going to have to tell us exactly what the problem statement is.
When you "tried to plug in 3x^2", what did it look like? Why were you discouraged and why did you do that?
If f(x) = 3x^2 - 3, then
f(2) = 3(2)^2 - 3
f(y) = 3(y)^2 - 3
f(z+2) = 3(z+2)^2 - 3
f(Houston) = 3(Houston)^2 - 3
When you "tried to plug in 3x^2", what did it look like? Why were you discouraged and why did you do that?
If f(x) = 3x^2 - 3, then
f(2) = 3(2)^2 - 3
f(y) = 3(y)^2 - 3
f(z+2) = 3(z+2)^2 - 3
f(Houston) = 3(Houston)^2 - 3
- Joined
- Feb 4, 2004
- Messages
- 16,582
I think the above means the following:
I was given the following function:
. . . . .\(\displaystyle f(x)\, =\, 3x^2\, -\, 3\)
Then I was told to plug this function into the following expression:
. . . . .\(\displaystyle \dfrac{f(x\, +\, \Delta x)\, -\, f(x)}{\Delta x}\)
I tried to plug just the "3x^2" into the expression, but I don't know what to do after that point.
Assuming that the above is correct, you are likely studying the topic of "function notation" or "evaluation of functions". To learn how to evaluate a function "at" an expression (rather than at a numerical value), please try here. After you have studied at least two lessons from the list, please attempt the exercise. Start with plugging the new input (x + delta-x) in where indicated. Your start of work should look something like the following:
. . . . .\(\displaystyle \dfrac{f(x\, +\, \Delta x)\, -\, f(x)}{\Delta x}\)
. . . . .\(\displaystyle \dfrac{\left[3(x\, +\, \Delta x)^2\, -\, 3\right]\, -\, [3x^2\, -\, 3]}{\Delta x}\)
...and so forth.
