How to find f'(0) from a left handed limit? (multiple questions)

[arbrelibre]

My homework is due really soon.
Here are all the questions I have absolutely NO idea how to do.


1.
Let:
f(X) = -6x^2+6x for x<0 and 7x^2-3 for x≥ 0

According to the definition of derivative, to computer f'(0), we need to compute the left hand limit
lim x-->0- =
and the right hand limit
lim x--> 0+
We conclude that f'(0)=

So...
I've figured out that the right hand limit is
[7x^2-3-(7(0)^2-3)]/(x-0)
and that f'(0)=DNE

My answer for the left hand limit is
[-6x^2+6x-(-6(0)^2+6(0))]/(x-0)
but the website won't accept my answer.

2.
Given the following table:

x----- 0.0097 ------- 0.0098 -------- 0.0099 -------- 0.01---- 0.0101 ----- 0.0101
f(x)-- 0.54783494--0.99814343 -- 0.46101272-- (-0.50636564)---- (-.9987636)

Calculate the value of f'(0.0099) to two place of accuracy.

3.
Let f(x) = 2/(x-8)
According to the definition of derivative, f'(x)= lim t-->x (2(x-8)-2(t-8))/((t-x)(t-8)(x-8))
The expression inside the limit simplifies to: 2/[-(x-8)/(t-8)]
Taking the limit of this fractional expression gives us
f′(x)= ?

Please, please, please help me. I am SO frustrated.

Thanks!

 

[haruspex]

My homework is due really soon.
Here are all the questions I have absolutely NO idea how to do.


1.
Let:
f(X) = -6x^2+6x for x<0 and 7x^2-3 for x≥ 0

According to the definition of derivative, to computer f'(0), we need to compute the left hand limit
lim x-->0- =
and the right hand limit
lim x--> 0+
We conclude that f'(0)=

So...
I've figured out that the right hand limit is
[7x^2-3-(7(0)^2-3)]/(x-0)
and that f'(0)=DNE

Wrong. Fill in the missing steps.

 

[arbrelibre]

Wrong. Fill in the missing steps.

What do you mean 'missing steps'?

 

[haruspex]

Let:
f(X) = -6x^2+6x for x<0 and 7x^2-3 for x≥ 0

According to the definition of derivative, to computer f'(0), we need to compute the left hand limit
lim x-->0- =
and the right hand limit
lim x--> 0+
We conclude that f'(0)=

So...
I've figured out that the right hand limit is
[7x^2-3-(7(0)^2-3)]/(x-0)
and that f'(0)=DNE

Pleae write out the steps between those last two statements so that I can see where you are going wrong.

 

[HallsofIvy]

My homework is due really soon.
Here are all the questions I have absolutely NO idea how to do.


1.
Let:
f(X) = -6x^2+6x for x<0 and 7x^2-3 for x≥ 0

According to the definition of derivative, to computer f'(0), we need to compute the left hand limit
lim x-->0- =
and the right hand limit
lim x--> 0+
We conclude that f'(0)=

So...
I've figured out that the right hand limit is
[7x^2-3-(7(0)^2-3)]/(x-0)
and that f'(0)=DNE

You are being far too "casual". What is "7x^2- 3- (7(0)^2- 3)" Actually write that out in detail.

My answer for the left hand limit is
[-6x^2+6x-(-6(0)^2+6(0))]/(x-0)
but the website won't accept my answer.

I don't see an answer! You haven't yet taken the limit.

2.
Given the following table:

x----- 0.0097 ------- 0.0098 -------- 0.0099 -------- 0.01---- 0.0101 ----- 0.0101
f(x)-- 0.54783494--0.99814343 -- 0.46101272-- (-0.50636564)---- (-.9987636)

Calculate the value of f'(0.0099) to two place of accuracy.

Okay, what have you done? Since you clearly know the formula, it's just a matter of arithmetic.

3.
Let f(x) = 2/(x-8)
According to the definition of derivative, f'(x)= lim t-->x (2(x-8)-2(t-8))/((t-x)(t-8)(x-8))

Where did you get that?

If f(x)=
The expression inside the limit simplifies to: 2/[-(x-8)/(t-8)]
Taking the limit of this fractional expression gives us
f′(x)= ?

As t goes to x, what is $\frac{x- 8}{t- 8}$?

Please, please, please help me. I am SO frustrated.

Thanks!