What is the derivative of f(x)=-x^3+4x^2 at (-1,5) using the limit definition?

  • Thread starter Precal_Chris
  • Start date
Oh, I see. You're trying to cancel out the h. That's not going to work.You have to do the expansion and then cancel out the h at the end.oh okso what do i do?Expand it out, and then try to cancel out the h.So far, you have:f'(x) = \lim_{h -> 0} \frac{-x^3-3x^2h+3xh^2+h^3+4x^2+8xh+4h^2+x^3-4x^2}{h}f'(x)= \lim_{h -> 0} \frac{h(-3x^2+3
  • #36
I don't know why people like long methods, when you have short ones...lol...
 
Physics news on Phys.org
  • #37
yes. now solve for your point.
 
  • #38
Precal_Chris said:
oh dang that's where i went wrong last time i put h as -1 ...-.-
im getting x and h confused...

so when h is taken out of the denominator i should have...

-3x(^2) -3xh -h(^2) +8x +4h

correct?
then placing 0's in for h should make it...

-3x(^2)- 3x(0)-(0)(^2) +8x + 4(0)...
leaving me with
-3x(^2)+8x...
right?

Bravo, now evaluate at x=-1.
 
  • #39
Precal_Chris said:
oh dang that's where i went wrong last time i put h as -1 ...-.-
im getting x and h confused...

so when h is taken out of the denominator i should have...

-3x(^2) -3xh -h(^2) +8x +4h

correct?
then placing 0's in for h should make it...

-3x(^2)- 3x(0)-(0)(^2) +8x + 4(0)...
leaving me with
-3x(^2)+8x...
right?

yep, you got it. Blahh, Snazzy was faster.
 
  • #40
sutupidmath said:
I don't know why people like long methods, when you have short ones...lol...

OP doesn't know the shortcut yet and was instructed to do this method.
 
  • #41
ok then plug in -1 for all the x's

which would be..

-3(-1)^2 +8(-1)

-3-8
-11?
 
  • #42
who's OP?
 
  • #43
looks good.
 
  • #44
Precal_Chris said:
who's OP?

You. OP=Original Post(er)
 
  • #45
oh...lol
our teacher said there was a way easier way to do this but she wouldn't tell us...
 
  • #46
yourdadonapogostick said:
OP doesn't know the shortcut yet and was instructed to do this method.

Well, with shortcut here i meant to use the other alternative of the deff. of the derivative,

[tex]\lim_{x\rightarrow -1}\frac{f(x)-f(-1)}{x+1}[/tex] it takes you much faster to the answer.
 
  • #47
Precal_Chris said:
oh...lol
our teacher said there was a way easier way to do this but she wouldn't tell us...

Well you will later learn the power rule, product rule, quotient rule, chain rule etc. and you will see that there is an easier way of doing these, but this was not my point, when i said: "i don't know why people like long methods when there are short ones".
 
  • #48
sutupidmath said:
Well, with shortcut here i meant to use the other alternative of the deff. of the derivative,

[tex]\lim_{x\rightarrow -1}\frac{f(x)-f(-1)}{x+1}[/tex] it takes you much faster to the answer.

I wouldn't call that a faster route. You end up with:

[tex]\lim_{x\rightarrow -1}\frac{-x^3+4x^2-5}{x+1}[/tex]
 
  • #49
Here's the short way ...

[tex]f(x)=-x^3+4x^2[/tex]

The Derivative ..

[tex]f'(-1)=-3x^2+8x=-3(-1)^2+8(-1)=-3-8=-11[/tex]

...

[tex]f(x)=ax^n[/tex]

[tex]f'(x)=anx^{n-1}[/tex]
 
Last edited:
  • #50
sutupidmath said:
"i don't know why people like long methods when there are short ones".
Maybe not but you usually act like a smart ass, but it's all good we still love you and we all do at one point in time :p
 
  • #51
Precal_Chris said:
oh...lol
our teacher said there was a way easier way to do this but she wouldn't tell us...

If your function is in the form of [itex]f(x)=ax^c[/itex], then your derivative is given by [itex]\frac{df}{dx}=cax^{c-1}[/itex] where a and c are constants. Since we know that [itex]\frac{d(f+g)}{dx}=\frac{df}{dx}+\frac{dg}{dx}[/itex], we can take the derivative of a polynomial VERY easily.
 
  • #52
wow and you guys knew how to do this all along...
thats crazy how do you guys know all of that?
 
  • #53
Because most, if not all of us, have done this stuff years ago.
 
  • #54
Precal_Chris said:
wow and you guys knew how to do this all along...
thats crazy how do you guys know all of that?
LOL, I was the same way last summer. Math is awesome :) If you can, take up to Calculus 3 and you'll be happy :)
 
  • #55
yeah at my high school they only have precal, cal AB, cal BC and then stats and stuff

but since i only have 2 more years of math left at high school this year I am taking precal
junior year I am taking ap stats
and senior year I am taking discrete math.. (easy math class but teaches you how to do real life stuff like balance check books and all that :/)
 
  • #56
Precal_Chris said:
oh...lol
our teacher said there was a way easier way to do this but she wouldn't tell us...

The shorter way is essentially an application of something called the power rule:

If [tex] f(x) = x^n then f'(x) = nx^{n-1}[/tex]

Wow, there are too many people trying to help :(
 
Last edited:
  • #57
Precal_Chris said:
wow and you guys knew how to do this all along...
thats crazy how do you guys know all of that?

Well, you will learn this stuff, don't worry about that. I didn't know you were still in high school...lol...
 
  • #58
Feldoh, using words in latex looks messy.
 
  • #59
yourdadonapogostick said:
Feldoh, using words in latex looks messy.


There is a way to write properly in LATEX, but i have no idea how, i also would be interested to know how to do it.
 
  • #60
rocomath said:
Maybe not but you usually act like a smart ass, but it's all good we still love you and we all do at one point in time :p

Yeah, probbably that's true what you are saying. I'll have to look upon that...lol..

Thanks for reminding me!That's what ( i do not know how to say this) friends are for.

P.S. This is not supposed to be a joke. I really meant this.
 
  • #61
sutupidmath said:
Well, with shortcut here i meant to use the other alternative of the deff. of the derivative,

[tex]\lim_{x\rightarrow -1}\frac{f(x)-f(-1)}{x+1}[/tex] it takes you much faster to the answer.

I'd prefer not to resort to synthetic division.
 
  • #62
yourdadonapogostick said:
I'd prefer not to resort to synthetic division.

thats what i did at first...
and i got the -x^3+4x^2-5 to go into (x+1)(-x^2 +5x -5)/(x+1) and so the (x+1) canceled...
but when i told everyone that i got -x^2 +5x -5 for my answer they said that there was no x+1 in the numerator and i had to go through the other equation...
and i just realized if i would have plugged in the -1 for all the x's i would have gotten -11 baack in like post number 15 XD
 
  • #63
sutupidmath said:
Well, you will learn this stuff, don't worry about that. I didn't know you were still in high school...lol...

is this for like mostly college people or something?
 
  • #64
Precal_Chris said:
is this for like mostly college people or something?


Well, one usually learns this stuff in calc 1. So, i wrongly guessed...lol..
 
  • #65
Precal_Chris said:
is this for like mostly college people or something?

I learned calculus in high school and I never went to college. I did join the military as a Naval Nuclear Mechanic, though.
 
  • #66
oh yeah she said this last six weeks we are going to be doing a lot of calc stuff
 

Similar threads

  • Precalculus Mathematics Homework Help
Replies
21
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
507
  • Precalculus Mathematics Homework Help
Replies
8
Views
621
  • Precalculus Mathematics Homework Help
Replies
25
Views
722
  • Precalculus Mathematics Homework Help
Replies
10
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
3K
  • Precalculus Mathematics Homework Help
Replies
4
Views
846
  • Precalculus Mathematics Homework Help
Replies
11
Views
967
  • Precalculus Mathematics Homework Help
Replies
11
Views
318
Back
Top