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.

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)

ow 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:

y ≈ (-10/19)*(2.8 - 3) + 1 = 1 2/19 ≈ 1.105

STEP-BY-STEP EXPLANATION:

y = (2x+13)/(2x^2+1)

y' = ((2x^2+1)*(2) - (2x+13)(4x)) / (2x^2+1)^2

at x=3, this is

y'(3) = (19*2 - 19*12)/19^2 = -10/19

So, your linearization is

y ≈ (-10/19)*(x-3) + 1

At x=2.8, this is

y ≈ (-10/19)*(2.8 - 3) + 1 = 1 2/19 ≈ 1.105

the final function for the graph becomes x-2=-1. before graphing you solve the equation, where x=1. then graph accordingly.   ( on the x-axis, 1 on the line will possess a vertical line.)

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: