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erastovalidia
10 days ago
11

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.

...?
Mathematics
2 answers:
lawyer [12K]10 days ago
6 0

RESULT:

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

DETAILED EXPLANATION:

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

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

When x=3, we get

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

Thus, the linear approximation is

y ≈ (-10/19)*(x-3) + 1

For x=2.8, we find

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

Inessa [12.1K]10 days ago
3 0

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)

<span>Now, we can apply this in a linear equation to determine a slope. Ty = -5x/8 +5(3)/8 +8/8 = -5x/8 +(15+8)/8 = -5x/8 +23/8 This will give us an estimated value at x=2.8 now</span>

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