A few quick questions with derivatives using limit def.

[Precal_Chris]

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)

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

The Attempt at a Solution

i don't even know where to begin ...but any help at all is welcomed and thanked.

 

[sutupidmath]

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)

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

The Attempt at a Solution

i don't 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:

$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h},f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$$


For your first function it would be

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

Second one:

$$f'(-1)=\lim_{x\rightarrow -1}\frac{-x^{3}+4x^{2} -f(-1)}{x-(-1)}$$

Third one

$$slope=m=f'(-1)=\lim_{x\rightarrow -1}\frac{\sqrt{x+10}-\sqrt{-1+10}}{x-(-1)}$$

Now all you need to do is evaluate those limits. DO u know how to do it?

 

[sutupidmath]

Blah, i forgot: You NEED to show your work next time, before anyone here can help you.

 

[Snazzy]

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:

$$\lim_{\substack{h\rightarrow 0}}\frac{f(x+h)-f(x)}{h}=\lim_{\substack{h\rightarrow 0}}\frac{2(x+h)^2+5-(2x^2+5)}{h}$$


$$=\lim_{\substack{h\rightarrow 0}}\frac{2x^2+4xh+2h^2+5-2x^2-5}{h}=\lim_{\substack{h\rightarrow 0}}\frac{4xh+2h^2}{h}$$


$$=\lim_{\substack{h\rightarrow 0}}4x+2h=4x$$

See how f(x) = 2x^2 + 5 and f(x+h) = 2(x+h)^2 + 5?

 

[sutupidmath]

To the OP:In order to receive 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.!

 

[Precal_Chris]

ok i didnt know that