Caitlin Esparza

2022-02-12

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, local minimum, or saddle point. Confirm your results using a graphing utility.

$f(x,\text{}{y}_{=}{(4x-1)}^{2}+{(2y+4)}^{2}+1$

zerogirlg16

Beginner2022-02-13Added 10 answers

Step 1

Differentiate$f(x,y)={(4x-1)}^{2}+{(2y+4)}^{2}+1$ partially with respect to x and y to obtain the values of $f}_{x$ and $f}_{y$ .

$f(x,y)={(4x-1)}^{2}+{(2y+4)}^{2}+1$

$f}_{x}=\frac{\partial {(4x-1)}^{2}+{(2y+4)}^{2}+1}{\partial x$

$=32x-8$

$f}_{y}=\frac{\partial {(4x-1)}^{2}+{(2y+4)}^{2}+1}{\partial y$

$=8(y+2)$

Step 2

Equate$f}_{x$ and $f}_{y$ to 0 and solve for x and y.

${f}_{x}=0$

$32x-8=0$

$x=\frac{8}{32}$

$=\frac{1}{4}$

${f}_{y}=8(y+2)$

$0=8(y+2)$

$y=-2$

The critical point is$(\frac{1}{4},\text{}-2)$ .

Step 3

Differentiate$f}_{x$ and $f}_{y$ partially with the respect to x and y respectively to obtain the value of $f}_{xx$ and $f}_{yy$ and differentiate $f}_{x$ with respect to y to obtain the value of $f}_{xy$ equate $f}_{xx$ and $f}_{yy$ to A and C respectively and $f}_{xy$ to B.

$f}_{xx}=\frac{\partial (32x-8)}{\partial x$

$=32$

$A=32$

$f}_{yy}=\frac{\partial \left(8(y+2)\right)}{\partial y$

$=8$

$C=8$

$f}_{xy}=\frac{\partial (32x-8)}{\partial y$

$=0$

$B=0$

Step 4

Substitute the values of A,B,C in$AC-{B}^{2}$ and obtain the value of the expression.

$AC-{B}^{2}=\left(32\right)\left(8\right)-{0}^{2}$

Differentiate

Step 2

Equate

The critical point is

Step 3

Differentiate

Step 4

Substitute the values of A,B,C in