How can I find the slope of a curve using the limit definition of a derivative?

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Problem statement
F(x)= x^2-3x
P=(1,f(1))

Revelant equation
Lim f(a+h)-f(a)/ (h) As h approaches 0

Attempt at a solution
ImageUploadedByPhysics Forums1393450782.190789.jpg


Now this is where I get stuck.. Because usually you have a h that cancels out and then you have something in the form of (h+a) that can be used to determine the slope.. What do I do next??
 
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grace77 said:
Problem statement
F(x)= x^2-3x
P=(1,f(1))

Revelant equation
Lim f(a+h)-f(a)/ (h) As h approaches 0

Attempt at a solution View attachment 67057

Now this is where I get stuck.. Because usually you have a h that cancels out and then you have something in the form of (h+a) that can be used to determine the slope.. What do I do next??
What you got for f(1 + h) is wrong. f(1 + h) = (1 + h)2 - 3(1 + h).

Also, you don't seem to use "=" consistently in your work. You have
##lim_{h \to 0} = ##, following by the thing you're taking the limit of, which is incorrect. You also don't connect the expressions you're doing algebra on with an equal sign, nor do you include the "lim" notation. "lim" should go away only when you actually take the limit, not before.
 
grace77 said:
Problem statement
F(x)= x^2-3x
P=(1,f(1))

Revelant equation
Lim f(a+h)-f(a)/ (h) As h approaches 0

Attempt at a solution View attachment 67057

Now this is where I get stuck.. Because usually you have a h that cancels out and then you have something in the form of (h+a) that can be used to determine the slope.. What do I do next??

You have made a calculation error:

F(1 + h) - F(1) = (1 + h)^2 - 3(1 + h) + 2
and not (1 + h)^2 + 2 as you have.
 
pasmith said:
You have made a calculation error:
F(1 + h) - F(1) = (1 + h)^2 - 3(1 + h) + 2

and not (1 + h)^2 + 2 as you have.
I see my error now thank you!
 
Mark44 said:
What you got for f(1 + h) is wrong. f(1 + h) = (1 + h)2 - 3(1 + h).

Also, you don't seem to use "=" consistently in your work. You have
##lim_{h \to 0} = ##, following by the thing you're taking the limit of, which is incorrect. You also don't connect the expressions you're doing algebra on with an equal sign, nor do you include the "lim" notation. "lim" should go away only when you actually take the limit, not before.
Thank you for your comments. How do I do the lim notations on physics forum?
 
grace77 said:
Thank you for your comments. How do I do the lim notations on physics forum?

Code:
[tex]\lim_{h \to 0}[/tex]
produces
<br /> \lim_{h \to 0}<br />
 
pasmith said:
Code:
[tex]\lim_{h \to 0}[/tex]

produces

<br /> <br /> \lim_{h \to 0}<br /> <br />
Is there a guide where I can find all of these notations?
 
grace77 said:
Is there a guide where I can find all of these notations?

See here.
 
pasmith said:
See here.
Thank you. You have been a great help!
 
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