Question

Verify the linear approximation at (0,0)

\(\frac{{2x + 3}}{{4y + 1}} \approx 3 + 2x - 12y\)

Short answer

The linearization of the function at that point is verified as \(\frac{{2x + 3}}{{4y + 1}} \approx 3 + 2x - 12y\)..

Step-by-step solution

Step-1(Equation used in finding the Linearization)

The equation use to find linearization is:-

\(L(x,y) = f(a,b) + {f_x}(a,b)(x - a) + {f_y}(a,b)(y - b)\),where \({f_x}\)and\({f_y}\) are partial derivatives..

Step-2(Calculation of Partial Derivative)

\(f(x,y) = \frac{{2x + 3}}{{4y + 1}}\)

\({f_x}(x,y) = \frac{2}{{4y + 1}}\)

\({f_x}(0,0) = \frac{2}{{4(0) + 1}} = 2\)

Now, \(f(x,y) = \frac{{2x + 3}}{{4y + 1}}\)

\({f_y}(x,y) = \frac{{ - 4(2x + 3)}}{{{{(4y + 1)}^2}}}\)

\(\)\({f_y}(0,0) = \frac{{ - 4(2(0) + 3)}}{{(4(0) + 1)}} = - 12\)

Partial Derivative exist at the point\((0,0)\) and are continuous at\((0,0)\)..

Therefore, Linearization is possible..

Step-3(Calculation of Linearization)

The function \(f(x,y) = \frac{{2x + 3}}{{4y + 1}}\) at the point \((0,0)\)is

\(f(0,0) = \frac{{2(0) + 3}}{{4(0) + 1}} = 3\)

\(L(x,y) = f(a,b) + {f_x}(a,b)(x - a) + {f_y}(a,b)(y - b)\)

\(L(x,y) = 3 + (x - 0)(2) + (y - 0)( - 12)\)

\( \approx 3 + 2x - 12y\)

Hence, The linearization of the function at point\((0,0)\) is\(3 + 2x - 12y\)..