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

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To find the derivative of the function f(x) = -x^3 + 4x^2 at the point (-1, 5) using the limit definition, the correct approach involves evaluating f(-1) first, which is 5. The limit definition for the derivative is f'(-1) = lim (x -> -1) [(f(x) - f(-1)) / (x + 1)]. After substituting and simplifying, the limit leads to the expression -3x^2 + 8x. Evaluating this at x = -1 gives a final derivative value of -11. The discussion highlights the importance of correctly applying the limit definition and algebraic manipulation to arrive at the solution.
  • #31
ok so i did the whole
-(x+h)^3
and the
4(x+h)^2
and the
+x^3-4x^2
and when i put them all together i got this:

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

all over H
this is after i canceled out the two x^3 and the two 4x^2's
 
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  • #32
so, cancel out the h. and then take the limit as h approaches 0.
 
  • #33
Feldoh meant as \lim_{h\rightarrow 0}

Now make h go to 0 and evaluate at x = -1.
 
  • #34
rocomath said:
Feldoh meant as \lim_{h\rightarrow 0}

Now make h go to 0 and evaluate at x = -1.

Yeah sorry it's \lim_{h-> 0}
 
  • #35
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?
 
  • #36
I don't know why people like long methods, when you have short ones...lol...
 
  • #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,

\lim_{x\rightarrow -1}\frac{f(x)-f(-1)}{x+1} 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,

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

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

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

f(x)=-x^3+4x^2

The Derivative ..

f'(-1)=-3x^2+8x=-3(-1)^2+8(-1)=-3-8=-11

...

f(x)=ax^n

f'(x)=anx^{n-1}
 
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 f(x)=ax^c, then your derivative is given by \frac{df}{dx}=cax^{c-1} where a and c are constants. Since we know that \frac{d(f+g)}{dx}=\frac{df}{dx}+\frac{dg}{dx}, 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 f(x) = x^n then f'(x) = nx^{n-1}

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.
 

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