Show the limit in a derivative.

Limits, differentiation, related rates, integration, trig integrals, etc.
GreenLantern
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Show the limit in a derivative.

Problem: Use the definition of f'(x) in terms to show that the derivative of $f(x)=3x^2-x-1$ is $f'(x)=6x-1$

My understanding is that I'm supposed to use the notion of the change in f(x) (also known as f'(x), correct?) to show a limit. As far as I'm concerned though, f(x) is just a parabola and doesn't ever experience the "limit" aspect anywhere so defining f'(x) as a limit seems a little silly.

Martingale
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Re: Show the limit in a derivative.

Problem: Use the definition of f'(x) in terms to show that the derivative of $f(x)=3x^2-x-1$ is $f'(x)=6x-1$

My understanding is that I'm supposed to use the notion of the change in f(x) (also known as f'(x), correct?) to show a limit. As far as I'm concerned though, f(x) is just a parabola and doesn't ever experience the "limit" aspect anywhere so defining f'(x) as a limit seems a little silly.

you are supposed to show that

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=6x-1$

for $f(x)=3x^2-x-1$

GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
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Re: Show the limit in a derivative.

Okay, so I'm still trying to wrap my head around this one, but I think I understand enough to make another post.

What we have normally dealt with is a limit being super super simple, like
$\lim_{h\to 1}$
To show that Y never actually is 1, but is always getting closer to it.

Now, the problem is asking me to show how Y is always approaching 6x-1 but never actually is.

Alright... I can do that.
$f(x)=3x^2-x-1$
$f'(x)=6x-1$
Now I make a little chart to show values to compare them.
When x=1
$f(1)=1$
$f'(1)=5$
When x=10
$f(1)=289$
$f'(1)=59$
But now I am immediately confused, because while this is normally how we showed that this was where the function never crossed we can see that it clearly does. Meaning that there is a point where f(x)=f'(x) and it's between 1&10 (I didn't take the time to find the exact point). How is this a "limit" if it does in fact cross over?

I'm not really sure where the
$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
is coming from. I'm assuming that's the general equation for a derivative?
So do I just plug all that in and solve for 6x-1?
What is h supposed to equal? 6x-1?
$\lim_{h\to 0}\frac{(3(6x-1)^2-(6x-1)-1))-(3x^2-x-1)}{6x-1}$
I don't have time at the moment, but I'll be back in an hour and solve this one...
Last edited by GreenLantern on Wed Mar 02, 2011 2:54 pm, edited 1 time in total.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

Re: Show the limit in a derivative.

Okay, so I'm still trying to wrap my head around this one, but I think I understand enough to make another post.

What we have normally dealt with is a limit being super super simple, like
$\lim_{h\to 1}$
To show that Y never actually is 1, but is always getting closer to it.

Now, the problem is asking me to show how Y is always approaching 6x-1 but never actually is.

Alright... I can do that.
$f(x)=3x^2-x-1$
$f'(x)=6x-1$
Now I make a little chart to show values to compare them.
When x=1
$f(1)=1$
$f'(1)=5$
When x=10
$f(1)=289$
$f'(1)=59$
But now I am immediately confused, because while this is normally how we showed that this was where the function never crossed we can see that it clearly does. Meaning that there is a point where f(x)=f'(x) and it's between 1&10 (I didn't take the time to find the exact point). How is this a "limit" if it does in fact cross over?

I have No idea what you are doing....

$\frac{d}{dx}\left[3x^2-x-1\right]=6x-1$

using the definition of a derivative

ie

show that

$\lim_{h\to 0}\frac{3(x+h)^2-(x+h)-1-(3x^2-x-1)}{h}=6x-1$

you might want to consult your book to remind yourself what a limit really is and what the definition of a derivative is.

GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
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Re: Show the limit in a derivative.

Whoa, okay. Fast posting. I did a little editing...

I'll be back, I have class. See if I can't get this all straightened out... Ya, I hit my book again before I come back though.