Consider the linear vector field i = azr, te ron . where a is an n x n constant matrix. suppose all the eigenvalues of a have negative real parts. then prove that x = is an asymptotically stable fixed point for this linear vector field. (hint: utilize a linear transformation of the coordinates which transforms a into jordan canonical form.)

the answer is c

hope this !

x=half

step-by-step explanation:

i think is g(1)=-1.

hope that i can

let f (x ) is a function and c be a positive real number,   the graph of the function g (x ) = f (cx ) is represented as follows.

when, c > 1,   the function compresses it in the x - direction.

when, 0 < c < 1, the function stretches it.

the function is g ( x ) = (9x )2.

the heighest power of x in g ( x ) = (9x )2 is 2, so the parent function is square function.

the equation of parent function is f ( x ) = x 2 .

the function is g ( x ) = (9x )2 and the parent function is f ( x ) = x 2 .

so, the functions f and g have the following relationship.

g ( x ) = f ( 9x ).

compare the function g ( x ) = f ( 9x ) with g (x ) = f (cx ).

c = 9 ( > 1 ).

so, the function g (x ) = (9x )2 compresses it in the x - direction.

step-by-step explanation: