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Mama L [17]
3 years ago
7

Find the local maximum and minimum values and saddle point(s) of the function.

Mathematics
1 answer:
iogann1982 [59]3 years ago
4 0
f(x,y)=9e^y(y^2-x^2)
\nabla f=\left\langle-18xe^y,9ye^y(2+y)\right\rangle

Critical points occur where the gradient is zero. This is guaranteed whenever x=0 and either y=0 or y=-2.

The Hessian matrix for this function looks like

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-18e^y&-18xe^y\\-18xe^y&9e^y(2-x^2+4y+y^2)\end{bmatrix}

and has determinant

|H(x,y)|=-162e^{2y}(2+x^2+4y+y^2)

Maxima occur whenever the determinant is positive and f_{xx}. Minima occur whenever both the determinant and f_{xx} are positive. Saddle points occur whenever the determinant is negative.

At (0,0), you have a saddle point since the determinant reduces to -324, so (0,0,0) is the saddle point.

At (0,-2), the determinant is \dfrac{324}{e^4}>0 and f_{xx}(0,-2)=-\dfrac{18}{e^2}, so \left(0,-2,\dfrac{36}{e^2}\right) is a local maximum.

No other critical points remain, so you're done.
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Martha wrote an example of a quadratic function for a homework assignment. The function she wrote is shown. f(x) = 5x3 + 2x2 + 7
Wittaler [7]

Answer:

The first term, 5x^3, can be eliminated.

The exponent on the first term, 5x^3, can be changed to a 2 and then combined with the second term, 2x^2

Step-by-step explanation:

The highest degree allowed in a quadratic function is 2, so the third degree term (the first term) needs to be eliminated or changed. The change shown above is one of many possibilities.

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For which function is f(x) equal to f^1(x) ( answer choices in picture )
Sonja [21]

Answer:

C. f(x)=\frac{x+1}{x-1}

Step-by-step explanation:

Let's find the inverse of each of the given options.

Option A:

f(x)=\frac{x+6}{x-6}\\y=\frac{x+6}{x-6}

To find f^{-1}(x), replace 'x' with 'y' and 'y' with 'x'. This gives,

x=\frac{y+6}{y-6}

Rewrite in terms of 'y'. This gives,

x(y-6)=y+6\\xy-6x=y+6\\xy-y=6x+6\\y=\frac{6x+6}{x-1}

The given function y=\frac{6x+6}{x-1}\ne y=\frac{x+6}{x-6}

So, option A is incorrect.

Option B:

f(x)=\frac{x+2}{x-2}\\y=\frac{x+2}{x-2}

To find f^{-1}(x), replace 'x' with 'y' and 'y' with 'x'. This gives,

x=\frac{y+2}{y-2}

Rewrite in terms of 'y'. This gives,

x(y-2)=y+2\\xy-2x=y+2\\xy-y=2x+2\\y=\frac{2x+2}{x-1}

The given function y=\frac{2x+2}{x-1}\ne y=\frac{x+2}{x-2}

So, option B is incorrect.

Option C:

f(x)=\frac{x+1}{x-1}\\y=\frac{x+1}{x-1}

To find f^{-1}(x), replace 'x' with 'y' and 'y' with 'x'. This gives,

x=\frac{y+1}{y-1}

Rewrite in terms of 'y'. This gives,

x(y-1)=y+1\\xy-x=y+1\\xy-y=x+1\\y=\frac{x+1}{x-1}

The given function y=\frac{x+1}{x-1}\ equals\ y=\frac{x+1}{x-1}

So, option C is correct.

Option D:

f(x)=\frac{x+5}{x-5}\\y=\frac{x+5}{x-5}

To find f^{-1}(x), replace 'x' with 'y' and 'y' with 'x'. This gives,

x=\frac{y+5}{y-5}

Rewrite in terms of 'y'. This gives,

x(y-5)=y+5\\xy-5x=y+5\\xy-y=5x+5\\y=\frac{5x+5}{x-1}

The given function y=\frac{5x+5}{x-1}\ne y=\frac{x+6}{x-6}

So, option D is incorrect.

Therefore, only option C is correct.

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Answer:

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Step-by-step explanation:

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