Given the following polynomial find the maximum possible number of turning points. f(x)=x^11+x^10+x^12+x^9
1 answer:
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Answer:
a) (8,8,-6)
b) 4x+4y+3z = -3
Step-by-step explanation:
a)
The surface is given by the equation
f(x,y,z) = 0 where
The gradient of this function is the vector
If we evaluate it in the point P = (-2,2,1) we obtain the point
(8,8,-6)
b)
The vectors with their tails at P are of the form
(-2,2,1)-(x,y,z) = (-2-x, 2-y, 1-z)
as they must be orthogonal to the gradient, they must be orthogonal to the vector (8,8,6) so their inner product is 0
and the equation of the desired plane is
4x+4y+3z = -3