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Find the value of f(1,3) for the function f(x,y)=2x²y+3y-5 - PowerPoint PPT Presentation

Find the value of f(1,3) for the function f(x,y)=2x²y+3y-5. 0 10 16 58. Evaluate f(x,y,z,t)=x³-4y²t+2zx when x=t=2, y=1, z=4. 4 8 16 20. Find for f(x,y)=2x³-y²+5xy-3. 2y²+5x 6x²-y²+5y 6x²+5y 6x²+5xy-3. Find for f(x,y)=3cos2x – 2sin3y. -6cosy -6cos3y -2cos3y

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Find the value of f(1,3) for the function f(x,y)=2x²y+3y-5

0

10

16

58

Evaluate f(x,y,z,t)=x³-4y²t+2zx when x=t=2, y=1, z=4

4

8

16

20

Find for f(x,y)=2x³-y²+5xy-3

2y²+5x

6x²-y²+5y

6x²+5y

6x²+5xy-3

Find for f(x,y)=3cos2x – 2sin3y

-6cosy

-6cos3y

-2cos3y

-2cosy

Find for f(x,y)=x³y²-4x+6y-9

2x³y - 4x + 6

2x³y + 6

2x³ + 6

2x³

Find for f(x,y)=y²-3x²-8y³x+1

-24y²

-6-24y²

-6x-24y

2-48y

Is (0,1) a stationary point for the function f(x,y)=5x²-8y²+3y-2?

Yes

No

Don’t know

Find the x coordinates of the stationary points of f(x,y)=x³-y²-3x+6y-8?

1 and 3

1 and -1

-1 and 3

-3 and -3

None of the above

Given D=fxx fyy-(fxy)², then which of the following below implies a stationary point is a local minimum?

1

2

3

4

1.

2.

3.

4.

Locate and determine the nature of the stationary points of f(x,y)=4x2-2y2+4y3-10

(0,0) saddle point, (0,1/3) local minimum

(0,0) saddle point, (0,-1/3) local minimum

None of the above

Given δf is the change in f at (x0,y0) resulting from small changes h, k to x0, y0 respectively and δf = f(x0+h, y0+k) - f(x0,y0).Then which of the following represents the relative error in f ?

1

2

3

4

5

1.

2.

3.

4.

None of the above

5.

Don’t know

Estimate the absolute error for the function f(x,y)=2x²y-5y²

δf ≈ 4xyδx + (2x²-10y)δy

δf ≈ 4xyδy + (2x²-10y)δx

δf ≈ 2x²δx + (2y-5y²)δy

δf ≈ 2x²δy + (2y-5y²)δx

Don’t know

Estimate the absolute error in f(x,y)=3x²-y²+xy-2 at the point x=2, y=4 if δx=±0.03 and δy=±0.02

± 0.78

± 0.52

± 0.36

± 0.2

If and x, y, u are subject to percentage relative errors of 2%, -3% and 1% respectively find the approximate percentage relative error in f.

-12%

-6%

-1%

2%

3%

Don’t know