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.
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Homework Statement


Use the limit definition to find the derivative at the indicated point.
f(x)= -x^3+4x^2, at (-1,5)


Homework Equations


f(a)=lim x->1 (f(x)-f(a))/(x-(a))


The Attempt at a Solution



ok so i got:
f'(-1)=(lim x->1) ((-x^3)+(4x^2)- f(-1))/(x-(-1))

f(-1)= (lim x->1) ((-x^3)+(4x^2)- f(-1))/(x+1)

my only question is .. what would f(-1) be..?

is this right?:

(1^3)-(4^2)
 
Last edited:
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No, incorrect. f(-1) is your function evaluated at -1. What is your function?
 
Last edited:
Precal_Chris said:

Homework Statement


Use the limit definition to find the derivative at the indicated point.
f(x)= -x^3+4x^2, at (-1,5)


Homework Equations


f(a)=lim x->1 (f(x)-f(a))/(x-(a))


The Attempt at a Solution



ok so i got:
f(-1)=(lim x->1) ((-x^3)+(4x^2)- f(-1))/(x-(-1))
f(-1)= (lim x->1) ((-x^3)+(4x^2)- f(-1))/(x+1)
You mean f '(-1) or df(-1)/dx, not f(-1), on the left.

my only question is .. what would f(-1) be..?

is this right?:

(1^3)-(4^2)
Since f(x)= -x3+ 4x2, f(-1)= -(-1)3+ 4(-1)2. What is that? (the "4" is NOT squared, only x= -1.)
 
ok so ultimately it would be..
-1+ 4
3..?
 
Precal_Chris said:
ok so ultimately it would be..
-1+ 4
3..?
Yes, so now put it all together and what is your limit?
 
ok

lim x->1 ((-x^3)+(4x^2)-3)/(x+1)

but idk what to do next because if i input 1 for all the x's then id get
0/2
 
i was thinking i could just factor out the
-x^3+4x^2-3 but when i do i don't get a (x+1) for one of my factors...
 
There is no such :''Last question"
 
lol.
 
  • #10
so does neone know how to help me from here?

(-x^3 + 4x^2 -3)/ (x+1)

becuz i don't think that's the final answer.

also i tried to make -1 one of the zeros..by using synthetic division
and that doesn't make a zero so i don't know how to cancel out the denominator...-.-

eeek please help someone :)
 
  • #11
If you're trying to find the derivative, your equation is wrong.
\frac{df}{dx}={\lim_{h\rightarrow{0}}}\frac{f(x+h)-f(x)}{h}
 
  • #12
but when I am givin a point then there's another formula you can use right?
 
  • #13
Find the derivative using that formula(there is an incredibly easy shortcut, but you were told to use the formula) and then plug in your point.
 
  • #14
ok ill try
but for the f(x+h)

would it be
-x^3-h^3+4x^2+4h^2?
and the for the -f(x)
of course it would be
-(-x^3 + 4x^2)

then you would get -h^3+4h^2/(h)
you could bring out an h
youd have

h(-h^2+4h)/h
which would cancel out the 2 h's
and youd be left with
-h^2+4h

and then plug them in?
 
  • #15
No. You plug (x+h) in for x when you are finding f(x+h).
 
  • #16
oh so it would be -(x+h)^3 + 4(x+h)^2 ?
 
  • #17
yes. That would be f(x+h)
 
  • #18
Precal_Chris said:
ok so ultimately it would be..
-1+ 4
3..?
No, that is NOT correct. Your function is f(x)= -x3+ 4x2.
f(-1)= -(-1)3+ 4(-1)2= 1+ 4= 5.
 
  • #19
so what do i do then?
im so confused:?:
like what equation do i need to use..and everything
 
  • #20
ok so its not 3 its 5...
that changes everything...

-x^3 +4x^2-5/(x+1)

then i can factor out an x+1
which the top and bottom x+1's cancel
leaving you with
-x^2+5x-5...

is this right so far? so i can go ahead and plug in for the x's?
 
  • #21
Where did the 5/(x+1) term come from? Go back to what I told you. I even did it myself; it works.
 
  • #22
Precal_Chris said:
ok so its not 3 its 5...
that changes everything...

-x^3 +4x^2-5/(x+1)

then i can factor out an x+1
which the top and bottom x+1's cancel
leaving you with
-x^2+5x-5...

is this right so far? so i can go ahead and plug in for the x's?

There is no top (x+1) term. Here, it is definitely easier to go about with this definition of the limit:

f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}

You'll have to deal with cubic powers, and square powers, but everything works out nicely at the end as long as you're careful with your algebra.

Using this definition of the limit:

f'(x)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}

Doesn't always work out.
 
  • #23
so:

-(x+h)^3 + 4(x+h)^2 - (-x^3+4x^2)/h

?

wouldnt that get me
-[(x+h)(x+h)(x+h)]
-[(x^2 + 2xh + h^2)(x+h)]
-[(x^3 + 3x^2h + 3xh^2 +h^3)]
(-x^3)-(3x^2h)-(3xh^2)-(h^3)
right?
 
  • #24
Your work is really ambiguous. I can't tell what you're doing or where you're expanding. Learn how to use LaTex.
 
  • #25
ok well atleast i got the other 2 right ill just try my best on this one because i don't seem to be getting newhere with it..
 
  • #26
alright i tried it the whole way through using that and i ended up getting

(-3x^2)+ 11x -5
 
  • #27
You made an algebra error somewhere.
 
  • #28
crap...
ok can you tell me really quick what you got so i can try it until i get it right?
ive been on the computer for over 2 hours now..
 
  • #29
f'(x) = \lim_{h -> 0} \frac{-(x+h)^3 + 4(x+h)^2 - (-x^3+4x^2)}{h}

Expand it out^^

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+3xh+h^2+8x+4h)}{h}
 
Last edited:
  • #30
Before you plug in your point, you will have a binomial.
 
  • #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
 
  • #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
 

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