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A few quick questions with derivatives using limit def.

1. Homework Statement
The first one says use the limit definition to find the derivative of the function.
F(x)= 1/(2x-4)

the second one is use the limit definition to find the derivative at the indicated point..
f(x)= -x^3 + 4x^2, at (-1,5)

the last one is use the limit process to find the slope of the graph of the function at the specified point.
f(x)= sqrt of (x + 10), at (-1,3)


2. Homework Equations
i know that this equation is used to help find it..
F(x + P)- f(x)/ P
p=the change in x..or delta x



3. The Attempt at a Solution
i dont even know where to begin ...but any help at all is welcomed and thanked.
 

Answers and Replies

1,631
4
1. Homework Statement
The first one says use the limit definition to find the derivative of the function.
F(x)= 1/(2x-4)

the second one is use the limit definition to find the derivative at the indicated point..
f(x)= -x^3 + 4x^2, at (-1,5)

the last one is use the limit process to find the slope of the graph of the function at the specified point.
f(x)= sqrt of (x + 10), at (-1,3)


2. Homework Equations
i know that this equation is used to help find it..
F(x + P)- f(x)/ P
p=the change in x..or delta x



3. The Attempt at a Solution
i dont even know where to begin ...but any help at all is welcomed and thanked.
Well, first of all the definition of the derivative, as you may know is:

[tex]f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h},f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}[/tex]


For your first function it would be

[tex]f'(x)=\lim_{h\rightarrow 0}\frac{\frac{1}{2(x+h)-4}-\frac{1}{2x-4}}{h}[/tex] Now you can go from here, right?

Second one:

[tex]f'(-1)=\lim_{x\rightarrow -1}\frac{-x^{3}+4x^{2} -f(-1)}{x-(-1)}[/tex]

Third one

[tex]slope=m=f'(-1)=\lim_{x\rightarrow -1}\frac{\sqrt{x+10}-\sqrt{-1+10}}{x-(-1)}[/tex]

Now all you need to do is evaluate those limits. DO u know how to do it?
 
1,631
4
Blah, i forgot: You NEED to show your work next time, before anyone here can help you.
 
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0
For example, if I wanted to find the derivative of the function 2x^2 + 5 using the definition of a limit, I would use the formula:

[tex]\lim_{\substack{h\rightarrow 0}}\frac{f(x+h)-f(x)}{h}=\lim_{\substack{h\rightarrow 0}}\frac{2(x+h)^2+5-(2x^2+5)}{h}[/tex]


[tex]=\lim_{\substack{h\rightarrow 0}}\frac{2x^2+4xh+2h^2+5-2x^2-5}{h}=\lim_{\substack{h\rightarrow 0}}\frac{4xh+2h^2}{h}[/tex]


[tex]=\lim_{\substack{h\rightarrow 0}}4x+2h=4x[/tex]

See how f(x) = 2x^2 + 5 and f(x+h) = 2(x+h)^2 + 5?
 
Last edited:
1,631
4
To the OP:In order to recieve further help, please show us what you have tried so far, and point out where are u stuck. Remember you are supposed to do your own homework not us.!!
 
ok i didnt know that
 

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