# Homework Help: Last question i swear

1. Mar 30, 2008

### Precal_Chris

1. The problem statement, all variables and given/known data
Use the limit definition to find the derivative at the indicated point.
f(x)= -x^3+4x^2, at (-1,5)

2. Relevant equations
f(a)=lim x->1 (f(x)-f(a))/(x-(a))

3. 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: Mar 30, 2008
2. Mar 30, 2008

### rocomath

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

Last edited: Mar 30, 2008
3. Mar 30, 2008

### HallsofIvy

You mean f '(-1) or df(-1)/dx, not f(-1), on the left.

Since f(x)= -x3+ 4x2, f(-1)= -(-1)3+ 4(-1)2. What is that? (the "4" is NOT squared, only x= -1.)

4. Mar 30, 2008

### Precal_Chris

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

5. Mar 30, 2008

### rocomath

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

6. Mar 30, 2008

### Precal_Chris

ok

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

but idk what to do next cuz if i input 1 for all the x's then id get
0/2

7. Mar 30, 2008

### Precal_Chris

i was thinking i could just factor out the
-x^3+4x^2-3 but when i do i dont get a (x+1) for one of my factors...

8. Mar 30, 2008

### sutupidmath

There is no such :''Last question"

9. Mar 30, 2008

### Precal_Chris

lol.

10. Mar 30, 2008

### Precal_Chris

so does neone know how to help me from here?

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

becuz i dont think thats the final answer.

also i tried to make -1 one of the zeros..by using synthetic division
and that doesnt make a zero so i dont know how to cancel out the denominator...-.-

eeek plz help someone :)

11. Mar 30, 2008

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. Mar 30, 2008

### Precal_Chris

but when im givin a point then theres another formula you can use right?

13. Mar 30, 2008

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. Mar 30, 2008

### 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?

15. Mar 30, 2008

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

16. Mar 30, 2008

### Precal_Chris

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

17. Mar 30, 2008

yes. That would be f(x+h)

18. Mar 30, 2008

### HallsofIvy

No, that is NOT correct. Your function is f(x)= -x3+ 4x2.
f(-1)= -(-1)3+ 4(-1)2= 1+ 4= 5.

19. Mar 30, 2008

### Precal_Chris

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

20. Mar 30, 2008

### 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?

21. Mar 30, 2008

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

22. Mar 30, 2008

### Snazzy

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. Mar 30, 2008

### 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?

24. Mar 30, 2008

### Snazzy

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

25. Mar 30, 2008

### Precal_Chris

ok well atleast i got the other 2 right ill just try my best on this one cuz i dont seem to be getting newhere with it..