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

$F(x) \approx F(x_0) + F'(x_0)(x-x_0)$

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

$2x^2y+y=2x+13$

Differentiating with respect to x, we have

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

Computing y' in the given point (3,1) we have

4(3)(1)+2(9)y'+y'=2

$y'=\frac{2-12}{19}$

$y'=-\frac{10}{19}$

The function will be approximated with the expression

$F(x) = 1 -\frac{10}{19}(x-3)$

To find the approximate value for x=2.8

$F(2.8) = 1-\frac{10}{19}(-0.2)=1.1$

The correct value is the option 1.1

6 0

3 years ago

Maggie creates the fraction models shown to find the value of 1 3/4 divided by 2/3

Assoli18 [71]

Answer: where is them model

Step-by-step explanation:

8 0

3 years ago