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

[Precal_Chris]

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)

 

[rocomath]

No, incorrect. f(-1) is your function evaluated at -1. What is your function?

 

[HallsofIvy]

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.)

 

[Precal_Chris]

ok so ultimately it would be..
-1+ 4
3..?

 

[rocomath]

ok so ultimately it would be..
-1+ 4
3..?

Yes, so now put it all together and what is your limit?

 

[Precal_Chris]

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

 

[Precal_Chris]

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...

 

[sutupidmath]

There is no such :''Last question"

 

[Precal_Chris]

lol.

 

[Precal_Chris]

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 :)

 

[yourdadonapogostick]

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}

 

[Precal_Chris]

but when I am givin a point then there's another formula you can use right?

 

[yourdadonapogostick]

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.

 

[Precal_Chris]

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?

 

[yourdadonapogostick]

No. You plug (x+h) in for x when you are finding f(x+h).

 

[Precal_Chris]

oh so it would be -(x+h)^3 + 4(x+h)^2 ?

 

[yourdadonapogostick]

yes. That would be f(x+h)

 

[HallsofIvy]

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.

 

[Precal_Chris]

so what do i do then?
im so confused:?:
like what equation do i need to use..and everything

 

[Precal_Chris]

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?

 

[yourdadonapogostick]

Where did the 5/(x+1) term come from? Go back to what I told you. I even did it myself; it works.

 

[Snazzy]

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.

 

[Precal_Chris]

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?

 

[Snazzy]

Your work is really ambiguous. I can't tell what you're doing or where you're expanding. Learn how to use LaTex.

 

[Precal_Chris]

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..

 

[Precal_Chris]

alright i tried it the whole way through using that and i ended up getting

(-3x^2)+ 11x -5

 

[Snazzy]

You made an algebra error somewhere.

 

[Precal_Chris]

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..

 

[Feldoh]

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}

 

[yourdadonapogostick]

Before you plug in your point, you will have a binomial.