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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #23 : Differentials

Find the total differential of the following function:

$f(x,y,z)=xz^4+y\sqrt{z}$

Possible Answers:

$z^4dx+\sqrt{z}dy+(4z^3+\frac{y}{2\sqrt{z}})dz$

$z^4dx+\sqrt{z}dy+(4xz^3+\frac{y}{2\sqrt{z}})dz$

$z^4+\sqrt{z}+4xz^3+\frac{y}{2\sqrt{z}}$

$z^4dx+\sqrt{z}dy+(4xz^3+\frac{y\sqrt{z}}{2})dz$

Correct answer:

$z^4dx+\sqrt{z}dy+(4xz^3+\frac{y}{2\sqrt{z}})dz$

Explanation:

The total differential of a function f(x,y,z) is given by

f_xdx+f_ydy+f_zdz

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=z^4 , $f_y=\sqrt{z}$ , $f_z=4xz^3+\frac{y}{2\sqrt{z}}$

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Example Question #24 : Differentials

Find the total differential of the function:

$f(x, y, z)=8z\ln(xy)$

Possible Answers:

$\frac{8z}{x}dx+\frac{8z}{y}dy+8\ln(xy)dz$

$\frac{8yz}{x}dx+\frac{8xz}{y}dy+8\ln(xy)dz$

$\frac{8z}{x}+\frac{8z}{y}+8\ln(xy)$

$\frac{8z}{y}dx+\frac{8z}{x}dy+8\ln(xy)dz$

Correct answer:

$\frac{8z}{x}dx+\frac{8z}{y}dy+8\ln(xy)dz$

Explanation:

The total differential of a function f(x,y,z) is given by

f_xdx+f_ydy+f_zdz

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=\frac{8z}{x}$ , $f_y=\frac{8z}{y}$ , $f_z=8\ln(xy)$

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Example Question #26 : Differentials

If $z=f(x,y)=x^6+3xy-\sin(xy)$ , calculate the total differential .

Possible Answers:

$dz=(6x^5+3y-y\cos(xy))dx+(3x-x\cos(xy))dy$

$dz=(6x^5-3y+y\cos(xy))dx+(3+x\cos(xy))dy$

$dz=(6x^5+3y-\cos(xy))dx+(3x-\cos(xy))dy$

$dz=(6x^5+3y+y\cos(xy))dx+(3x+x\cos(xy))dy$

$dz=(\frac{x^7}{7}+3y-y\cos(xy))dx+(\frac{3x^2}{2}-x\cos(xy))dy$

Correct answer:

$dz=(6x^5+3y-y\cos(xy))dx+(3x-x\cos(xy))dy$

Explanation:

The total differential of a function z=f(x,y) is defined as the sum of the partial derivatives of with respect to each of its variables; that is,

$dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

In this case, $z=f(x,y)=x^6+3xy-\sin(xy)$ , and so we use the sum rule, the rule for derivatives of a variable raised to a power, the rule for the derivative of $\sin(x)$ , and the chain rule to calculate the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ , as shown:

$\frac{\partial f}{\partial x}=6x^5+3y-y\cos(xy)$ ,

$\frac{\partial f}{\partial y}=3x-x\cos(xy)$ .

In both cases, we treated the variable not being differentiated as a constant, and applied the chain rule to $\sin(xy)$ to calculate its partial derivatives. Now that $\frac{\partial f}{\partial x}dx$ and $\frac{\partial f}{\partial y}dy$ have been calculated, all that remains is to substitute them into the definition of the total derivative:

$dz=(6x^5+3y-y\cos(xy))dx+(3x-x\cos(xy))dy$

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Example Question #27 : Differentials

If z=f(x,y)=xe^x^y , calculate the total differential .

Possible Answers:

e^x^y(1+xy)+2xe^x^y

e^x^y(1+xy)+x^2e^x^y

e^x^y(1+y)+x^2e^x^y

e^x^y(1+xy)+xe^x^y

e^x^y(1+x)+x^2e^x^y

Correct answer:

e^x^y(1+xy)+x^2e^x^y

Explanation:

The total differential of a function z=f(x,y) is defined as the sum of the partial derivatives of with respect to each of its variables; that is,

$dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

In this case, z=f(x,y)=xe^x^y , and so we use the product rule and the rule for differentiating e^x to calculate the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ , as shown:

$\frac{\partial f}{\partial x}=(1)(e^x^y)+(x)(ye^x^y)=e^x^y+xye^x^y=e^x^y(1+xy)$ ,

$\frac{\partial f}{\partial y}=(x)xe^x^y=x^2e^x^y$

In both cases, we treated the variable not being differentiated as a constant, and applied the product rule to calculate its partial derivatives. Now that $\frac{\partial f}{\partial x}dx$ and $\frac{\partial f}{\partial y}dy$ have been calculated, all that remains is to substitute them into the definition of the total derivative:

dz=e^x^y(1+xy)dx+(x^2e^x^y)dy

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Example Question #1447 : Partial Derivatives

Compute the differentials for the following function.

$z=e^{x^2+4y^2}\sin(y)$

Possible Answers:

$dz=8y\cos(y)e^{x^2+4y^2} dy$

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

$dz=2xe^{x^2+4y^2}\sin(y)+8y\cos(y)e^{x^2+4y^2}$

$dz=2xe^{x^2+4y^2}\sin(y)$

$dz=2xe^{x^2+4y^2}\sin(y)dx$

Correct answer:

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Explanation:

What we need to do is take derivatives, and remember the general equation.

$z=e^{x^2+4y^2}\sin(y)$

w=g(x,y,z)

dw=g_xdx+g_ydy+g_zdz

When taking the derivative with respect to y recall that the product rule needs to be used.

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

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Example Question #1441 : Partial Derivatives

Compute the differentials for the following function.

$z=e^{x^2+4y^2}\sin(y)$

Possible Answers:

$dz=2xe^{x^2+4y^2}\sin(y)$

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

$dz=2xe^{x^2+4y^2}\sin(y)dx$

$dz=8y\cos(y)e^{x^2+4y^2} dy$

$dz=2xe^{x^2+4y^2}\sin(y)+8y\cos(y)e^{x^2+4y^2}$

Correct answer:

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Explanation:

What we need to do is take derivatives, and remember the general equation.

$z=e^{x^2+4y^2}\sin(y)$

w=g(x,y,z)

dw=g_xdx+g_ydy+g_zdz

When taking the derivative with respect to y recall that the product rule needs to be used.

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Report an Error

Example Question #1 : Relative Minimums And Maximums

Find and classify all the critical points for f(x,y)=5x^2y+3y^2+2x^2-4y^3+1 .

Possible Answers:

(0,0): Relative Minimum

$(0,\frac{1}{2}):$ Saddle Point

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

(0,0): Relative Minimum

$(0,\frac{1}{2}):$ Saddle Point

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Relative Maximum

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Relative Minimum

(0,0): Relative Minimum

$(0,\frac{1}{2}):$ Relative Minimum

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

(0,0): Relative Maximum

$(0,\frac{1}{2}):$ Relative Minimum

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

(0,0): Saddle Point

$(0,\frac{1}{2}):$ Saddle Point

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

Correct answer:

(0,0): Relative Minimum

$(0,\frac{1}{2}):$ Saddle Point

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

Explanation:

First thing we need to do is take partial derivatives.

f_x(x,y)=10xy+4x

$f_{xx}(x,y)=10y+4$

f_y(x,y)=5x^2+6y-12y^2

$f_{yy}(x,y)=6-24y$

$f_{xy}(x,y)=10x$

Now we want to find critical points, we do this by setting the partial derivative in respect to x equal to zero.

$10xy+4x=0 \rightarrow 2x(5y+2)=0\rightarrow x=0, y=-\frac{2}{5}$

Now we want to plug in these values into the partial derivative in respect to y and set it equal to zero.

x=0:

$5(0)^2+6y-12y^2=0\rightarrow 6y(1-2y)=0\rightarrow y=0, y=\frac{1}{2}$

$y=-\frac{2}{5}:$

$5x^2+6(-\frac{2}{5})-12(\frac{2}{5})^2=0\rightarrow x^2=\frac{108}{125}\rightarrow x=\pm\sqrt{\frac{108}{125}}$

Lets summarize the critical points:

If x=0:

$(0,0), (0, \frac{1}{2})$

If $y=-\frac{2}{5}:$

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big), \Big(-\sqrt{\frac{108}{125}},- \frac{2}{5}\Big)$

Now we need to classify these points, we do this by creating a general formula .

$D=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^2$ , where a,b , is a critical point.

If D>0 and $f_{xx}(a,b)>0$ , then there is a relative minimum at (a,b)

If D>0 and $f_{xx}(a,b)<0$ , then there is a relative maximum at (a,b)

If D<0 , there is a saddle point at (a,b)

If D=0 then the point (a,b) may be a relative minimum, relative maximum or a saddle point.

D=(10y+4)(6-24y)-(10x)^2

Now we plug in the critical values into .

(0,0):

D=(10(0)+4)(6-24(0))-(10(0))^2=(4)(6)=24

$f_{xx}(0,0)=10(0)+4=4$

Since D>0 and $f_{xx}(a,b)>0$ , (0,0) is a relative minimum.

$(0,\frac{1}{2}):$

$D=(10(\frac{1}{2})+4)(6-24(\frac{1}{2}))-(10(0))^2=(5+4)(6-12)=-54$

Since D<0 , $(0,\frac{1}{2})$ is a saddle point.

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$

$D=(10(-\frac{2}{5})+4)(6-24(-\frac{2}{5}))-(10(\sqrt{\frac{108}{125}}))^2=(0)-\frac{432}{5}=-\frac{432}{5}$

Since D<0 , $\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big)$ is a saddle point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$

$D=(10(-\frac{2}{5})+4)(6-24(-\frac{2}{5}))-(10(-\sqrt{\frac{108}{125}}))^2=(0)-\frac{432}{5}=-\frac{432}{5}$

Since D<0 , $\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big)$ is a saddle point

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Example Question #2 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=5x^2+5y^2+4xy-3 .

Possible Answers:

(0,0) is a relative maximum.

(5,5) is a relative maximum.

(5,5) is a relative minimum.

(0,0) is a relative minimum.

Correct answer:

(0,0) is a relative minimum.

Explanation:

To find the relative maxima and minima, we must find all the first order and second order partial derivatives. We will use the first order partial derivative to find the critical points, then use the equation

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=10x+4y$

$\frac{\partial f }{\partial y}=f_y(x,y)=10y+4x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=10$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=10$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=4$

To find the critical points, we will set the first derivatives equal to

f_x(x,y)=10x+4y=0

f_y(x,y)=10y+4x=0

$10x+4y=0\Rightarrow 10x=-4y\Rightarrow x=-4y/10$

$10y+4x=0\Rightarrow 10y+4(-4y/10)=0 \Rightarrow 10y-16y/10=0$

84y/10=0

y=0

x=-4y/10=-4(0)/10=0

There is only one critical point and it is at (0,0) . We need to determine if this critical point is a maximum or minimum using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=10(10)-(4)^2=100-16=84

$f_{xx}(x,y)=10$

Since D(x,y)>0 and $f_{xx}(x,y)>0$ , (0,0) is a relative minimum.

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Example Question #3 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=x^3+4y^3-2xy+17 .

Possible Answers:

(0,0) is a relative maximum, $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ is a relative minimum

(0,0) is a saddle point, $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ is a relative minimum

(0,0) and $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ are relative minima

(0,0) and $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ are relative maxima

Correct answer:

(0,0) is a saddle point, $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ is a relative minimum

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=3x^2-2y$

$\frac{\partial f }{\partial y}=f_y(x,y)=12y^2-2x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=6x$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=24y$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=-2$

To find the critical points, we will set the first derivatives equal to

$f_x(x,y)=3x^2-2y=0\Rightarrow 2y=3x^2\Rightarrow y=3x^2/2$

f_y(x,y)=12y^2-2x=0

$f_y(x,y)=12(3x^2/2)^2-2x=0\Rightarrow 27x^4-2x=0$

x(27x^3-2)=0

There are two possible values of , x=0 and $x=\frac{\sqrt[3]{2}}{3}$ .

We find the corresponding values of using y=3x^2/2 (found by rearranging the first derivative)

y(0)=3(0)^2/2=0

$y=\frac{3\left ( \frac{\sqrt[3]{2}}{3} \right )^2}{2} =\frac{3\left ( \frac{\sqrt[3]{4}}{9} \right )}{2} =\frac{\sqrt[3]{4}}{6}$

There are critical points at (0,0) and $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ . We need to determine if the critical points are maximums or minimums using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$D(x,y)=(6x)(24y)-\left ( -2 \right )^2=144xy-4$

At (0,0) ,

D(x,y)=144(0)(0)-4=-4<0

Since D(x,y)<0 , (0,0) is a saddle point.

At $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ ,

$D(x,y)=144\left ( \frac{\sqrt[3]{2}}{3} \right )\left ( \frac{\sqrt[3]{4}}{6} \right )-4=144\left ( \frac{\sqrt[3]{8}}{18} \right )-4$

$=144\left ( \frac{2}{18} \right )-4=12>0$

$f_{xx}(x,y)=6\left ( \frac{\sqrt[3]{2}}{3} \right )=2\sqrt[3]{2}>0$

Since D(x,y)>0 and $f_{xx}(x,y)>0$ , $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ is a relative minimum.

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Example Question #4 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=x^2y+4y^3+xy+1 .

Possible Answers:

(0,0) , (-1,0) and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ are relative maxima.

(0,0) , (-1,0) and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ are saddle points.

(0,0) , (-1,0) and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ are relative minima.

(0,0) and (-1,0) are relative minima, $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ is a relative maximum.

Correct answer:

(0,0) , (-1,0) and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ are saddle points.

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=2xy+y$

$\frac{\partial f }{\partial y}=f_y(x,y)=x^2+12y^2+x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2y$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=24y$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=2x+1$

To find the critical points, we will set the first derivatives equal to

$f_y(x,y)=x^2-12y^2+x=0\Rightarrow 12y^2=-x^2-x$

$y=\sqrt{\frac{-x(x+1)}{12}}$

$f_x(x,y)=2xy+y=0\Rightarrow y(2x+1)=0$

$\sqrt{\frac{-x(x+1)}{12}}(2x+1)=0$

Squaring both sides of the equation gives us

$\frac{-x(x+1)}{12}(2x+1)^2=0^2$

Multiplying both sides of the equation by gives us

-x(x+1)(2x+1)^2=0

There are three possible values of ; x=0 , x=-1 and x=-1/2 .

We find the corresponding values of using $y=\sqrt{\frac{-x(x+1)}{12}}$ (found by rearranging the first derivative)

$y(0)=\sqrt{\frac{-(0)(0+1)}{12}}=\sqrt{\frac{0}{12}}=0$

$y(-1)=\sqrt{\frac{-(-1)(-1+1)}{12}}=\sqrt{\frac{-(-1)(0)}{12}}=0$

$y(-1/2)=\sqrt{\frac{-(-1/2)(-1/2+1)}{12}}=\sqrt{\frac{-(-1/2)(1/2)}{12}}$

$=\sqrt{\frac{(1/4)}{12}}=\sqrt{\frac{1}{48}}$

There are critical points at (0,0) , (-1,0) and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ . We need to determine if the critical points are maximums or minimums using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=(2y)(24y)-(2x+1)^2=48y^2-(4x^2+4x+1)

D(x,y)=48y^2-4x^2-4x-1

At (0,0) ,

D(x,y)=48(0)^2-4(0)^2-4(0)-1=-1<0

Since D(x,y)<0 , (0,0) is a saddle point.

At (-1,0) ,

D(x,y)=48(0)^2-4(-1)^2-4(-1)-1=0-4-4-1

D(x,y)=-9<0

Since D(x,y)<0 , (-1,0) is a saddle point.

At $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ ,

$D(x,y)=48\left ( \sqrt{\frac{1}{48}} \right )^2-4(1/2)^2-4(1/2)-1$

$D(x,y)=48\left ( \frac{1}{48} \right )-4(1/4)-4(1/2)-1$

D(x,y)=1-1-2-1=-3<0

Since D(x,y)<0 , $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ is a saddle point.

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Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

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You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

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* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

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Calculus 3 : Calculus 3

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Example Question #23 : Differentials

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