14. Find the absolute maximum and minimum values of the following function on the given region $\mathrm{R}$.
\[
f(x, y)=2 x^{2}-4 x+3 y^{2}+3 ; R=\left\{(x, y):(x-1)^{2}+y^{2} \leq 1\right\}
\]
Find both partial derivatives of $f(x, y)$.
\[
f_{x}(x, y)=\square \text { and } f_{y}(x, y)=
\]
Determine the absolute maximum value of $f$ on $\mathrm{R}$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The absolute maximum value of $f$ on $R$ is (Simplify your answer.)
B. There is no absolute maximum value.
Determine the absolute minimum value of $f$ on R. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The absolute minimum value of $f$ on $R$ is (Simplify your answer.)
B. There is no absolute minimum value.

Step 1 ：First, we find the partial derivatives of the function. The partial derivative of \(f(x, y)\) with respect to \(x\) is \(f_x(x, y) = 4x - 4\), and the partial derivative with respect to \(y\) is \(f_y(x, y) = 6y\).

Step 2 ：Next, we set these partial derivatives equal to zero to find the critical points. Solving \(f_x(x, y) = 0\) gives \(x = 1\), and solving \(f_y(x, y) = 0\) gives \(y = 0\). So the only critical point in the region \(R\) is \((1, 0)\).

Step 3 ：We evaluate the function at this critical point to get \(f(1, 0) = 2(1)^2 - 4(1) + 3(0)^2 + 3 = 1\).

Step 4 ：Next, we need to check the boundary of the region \(R\), which is the circle \((x - 1)^2 + y^2 = 1\). We substitute \(x = 1 + \cos \theta\) and \(y = \sin \theta\) into the function, where \(\theta\) ranges from \(0\) to \(2\pi\).

Step 5 ：The function becomes \(f(\theta) = 2(1 + \cos \theta)^2 - 4(1 + \cos \theta) + 3 \sin^2 \theta + 3\).

Step 6 ：We find the derivative of this function with respect to \(\theta\), set it equal to zero, and solve for \(\theta\) to find the maximum and minimum values on the boundary. The derivative is \(f'(\theta) = -4 \sin \theta (2 \cos \theta + 1) + 6 \sin \theta \cos \theta\).

Step 7 ：Setting this equal to zero gives \(\sin \theta = 0\) or \(\cos \theta = -1/2\). So the possible values of \(\theta\) are \(0\), \(\pi\), \(2\pi/3\), and \(4\pi/3\).

Step 8 ：We evaluate the function at these values of \(\theta\) to get \(f(0) = 1\), \(f(\pi) = 5\), \(f(2\pi/3) = 7/2 + 3\sqrt{3}/2\), and \(f(4\pi/3) = 7/2 - 3\sqrt{3}/2\).

Step 9 ：So the maximum value of the function on the region \(R\) is \(7/2 + 3\sqrt{3}/2\), and the minimum value is \(1\).

Step 10 ：Thus, the absolute maximum value of \(f\) on \(R\) is \(\boxed{7/2 + 3\sqrt{3}/2}\), and the absolute minimum value of \(f\) on \(R\) is \(\boxed{1}\).