# Find the critical points of the following functions.Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, a local minimum,or a saddle point. If the Second Derivative Test is inconclusive,determine the behavior of the function at the critical points. f(x,y)=(4x-1)^2+(2y+4)^2+1

Question
Analyzing functions
Find the critical points of the following functions.Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, a local minimum,or a saddle point. If the Second Derivative Test is inconclusive,determine the behavior of the function at the critical points.
$$\displaystyle{f{{\left({x},{y}\right)}}}={\left({4}{x}-{1}\right)}^{{2}}+{\left({2}{y}+{4}\right)}^{{2}}+{1}$$

2020-11-09
Given function $$\displaystyle{f{{\left({x},{y}\right)}}}={\left({4}{x}-{1}\right)}^{{2}}+{\left({2}{y}+{4}\right)}^{{2}}+{1}={16}{x}^{{2}}+{4}{y}{2}-{8}{x}+{16}{y}+{18}$$
Critical points:
$$\displaystyle{f}_{{x}}=\frac{{\partial{f}}}{{\partial{x}}}={32}{x}-{8}$$
$$\displaystyle{f}_{{y}}=\frac{{\partial{f}}}{{\partial{y}}}={8}{y}+{16}$$
Setting these equal to zero.
$$\displaystyle{32}{x}-{8}={0}\Rightarrow{x}=\frac{{1}}{{4}}$$
$$\displaystyle{8}{y}+{16}={0}\Rightarrow{y}=-{2}$$
so, the critical point is $$\displaystyle{\left(\frac{{1}}{{4}},-{2}\right)}.$$
Now find the second derivatives,
$$\displaystyle{f}_{{\times}}={32},{f}_{{{y}{y}}}={8},{f}_{{{x}{y}}}={0}$$
So, $$\displaystyle{D}={f}_{{\times}}\cdot{f}_{{{y}{y}}}-{\left({f}_{{{x}{y}}}\right)}^{{2}}$$
For $$\displaystyle{\left(\frac{{1}}{{4}},-{2}\right)},$$
$$\displaystyle{D}={32}\cdot{8}-{0}={256}{>}{0}$$
Since D>0 and $$\displaystyle{f}_{{\times}}={32}{>}{0}$$, so the obtaines critical point is local minimum.
The local minimum at $$\displaystyle{\left(\frac{{1}}{{4}},-{2}\right)}$$ is 1.

### Relevant Questions

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