Local linearization

1. Apr 7, 2005

UrbanXrisis

the question is here

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

so I need to find the formula of the graph.

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

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

where did I go wrong?

2. Apr 7, 2005

Hurkyl

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

3. Apr 7, 2005

UrbanXrisis

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

what I did was...

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

4. Apr 7, 2005

Hurkyl

Staff Emeritus
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.

5. Apr 7, 2005

UrbanXrisis

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

6. Apr 7, 2005