Question

The following curve passes through (3,1). Use the local linearization of the curve to find the approximate value of y at x=2.8.

Answer 1

the following? curve?

Answer 2

we need a curve in order to get the derivative, slope of the tangent line at 3,1 in order to approximate it with

Answer 3

can we do it without a curve? like what if it just said a curve passes through (3,1).....

Answer 4

no, becasue the linearization depends on the tangent line to the curve at that point; the point itself has no value to it alone

Answer 5

oh the curve is 2x^2y+y=2x+13

Answer 6

d/dx (2 x^2 y + y = 2x + 13)

4xy + 2x^2 y' + y' = 2

4xy + y'(2x^2 + 1) = 2

y' = (2- 4xy)/(2x^2 +1)

Answer 7

that is the equation for the slope of the tangent line, now we use 3,1 to get the slopes value

Answer 8

y' = (2- 4(3)(1))/(2(3)^2 +1)
= 2 - 12 / 18 +1
= -10/19 = -5/8

Answer 9

now we can use this in a linear equation for a slope

Ty = -5x/8 +5(3)/8 +8/8
= -5x/8 +(15+8)/8
= -5x/8 +23/8

this will gives us an approximation at x=2.8 now

Answer 10

okay thanks

Answer 11

youre welcome

Answer 12

But, these are my posibble answer??

0.5
0.9
0.95
1.1
1.4

Answer 13

http://www.wolframalpha.com/input/?i=-5%282.8%29%2F8+%2B23%2F8

1.125 is close to 1.1

Answer 14

got you