(820,#21) Use computer to find the critical points of correct to three decimal places. Then classify the critical points and find the highest and lowest points on the graph.

Solution: The equations for the critical points are

We use Maple to find the roots of the cubic:

evalf(solve(4*x^3-10*x+3=0,x));

1.401771248 + .3 10 ^(-9) I, -1.713939436, .3121681880 - .2 10^(-9) I

The "10^(-9) I " terms, where , are computer noise, round-off errors. (You can verify this by graphing 4*x^3-10*x+3 to see that it has three real roots)

The critical points are, to three decimal places: ( 1.402, 0), (-1.714, 0), (0.312, 0).

Next, we graph in the neighborhood of each critical point to determine its nature. 


plot3d([x,y,-(x^4-5*x^2+y^2+3*x+2)],x=-2.5..-0.5,y=-0.5..1,color=aquamarine,
             style =patch, axes = framed,title=`-f(x,y)`);

f(x,y) has a minimum at (-1.714,0). 


plot3d([x,y,x^4-5*x^2+y^2+3*x+2],x=0.2..0.5,y=-0.5..0.5,
              color=red ,style=patch ,axes = framed,title=`f(x,y)`);

f(x,y) has a saddle at (0.312,0). 


plot3d([x,y,-(x^4-5*x^2+y^2+3*x+2)],x=1..1.7,y=-0.5..0.5,color=green,
              style=patch, axes=framed,title=`-f(x,y)`);

f(x,y) has a minimum at (1.402,0)

To conclude, for large x and larbe y the function

will have large, positive values. By continuity, it will have to have a " lowest" point. Looking at our three graphs, we see that this miminum will occur at x = -1.714, y = 0, where the value is -9.200.

by:sh