Answer:
Option 1.1
Step-by-step explanation:
The linearization of a curve implies the use of calculus to find the local value for the derivative and approximating the function by the use of the formula
[tex]F(x) \approx F(x_0) + F'(x_0)(x-x_0)[/tex]
The function is given in such way that it's much easier to find the derivative by implicit differentiation than isolating any of the variables
[tex]2x^2y+y=2x+13[/tex]
Differentiating with respect to x, we have
[tex]4xy+2x^2y'+y'=2[/tex]
Computing y' in the given point (3,1) we have
4(3)(1)+2(9)y'+y'=2
[tex]y'=\frac{2-12}{19}[/tex]
[tex]y'=-\frac{10}{19}[/tex]
The function will be approximated with the expression
[tex]F(x) = 1 -\frac{10}{19}(x-3)[/tex]
To find the approximate value for x=2.8
[tex]F(2.8) = 1-\frac{10}{19}(-0.2)=1.1[/tex]
The correct value is the option 1.1
