Finding the linear approximation formula for a graph

[UrbanXrisis]

the question is http://home.earthlink.net/~urban-xrisis/clip001.jpg

I got a different answer than what the book says...

so I need to find the formula of the graph.

$$H'(3)=f(3)=2$$
$$m=\frac{\Delta y}{\Delta x}$$

$$2=\frac{\Delta y}{x- \int^3_0 f(t)dt}$$
$$y=2(x+2)$$
$$y=2x+4$$

the book's answer is 2x-8

where did I go wrong?

 

[Hurkyl]

Through what point did you want your line to go through? It looks like you used (-2, 0)...

 

[UrbanXrisis]

you mean I should do:
$$2=\frac{y-2}{x-3}$$
$$y=2x-4$$
??

what I did was...

$$H'(x)=\frac{y-\int_{-2}^yf(t)dt}{x-\int_0^xf(t)dt}$$

 

[Hurkyl]

Why do you want your line to go through the point (3, 2)?

What you need to do is stop guessing and think it through. Working through a simpler problem might help.

What is the local linearization of the function f(x) = x² near x = -1? First tell me what that means geometrically, then work out the answer algebraically.

 

[UrbanXrisis]

f'(x)= (y2-y1) / (x2-x1)
-2= (y2-1) / (x+1)
y=-2x-1

 

[UrbanXrisis]

thank you, I used your example to get the right answer

 

[Hurkyl]

I notice you didn't try a geometric explanation.

Anyways, that's exactly right. Now, why did you pick the point (x1, y1) = (-1, 1)? Apply the same reasoning to your problem.