9514 1404 393
Answer:
- (c1, c2, c3) = (-2t, 4t, t) . . . . for any value of t
- NOT linearly independent
Step-by-step explanation:
We want ...
c1·f1(x) +c2·f2(x) +c3·f3(x) = g(x) ≡ 0
Substituting for the fn function values, we have ...
c1·x +c2·x² +c3·(2x -4x²) ≡ 0
This resolves to two equations:
x(c1 +2c3) = 0
x²(c2 -4c3) = 0
These have an infinite set of solutions:
c1 = -2c3
c2 = 4c3
Then for any parameter t, including the "trivial" t=0, ...
(c1, c2, c3) = (-2t, 4t, t)
__
f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)
Answer:
g(x) = x is the base function.
To shrink it vertically by 1/2, make 1/3 the coefficient of the variable x.
To shift it 4 units to the right, subtract 4 from within the squared variable.
To shift 5 units down, subtract 5 from the entire function.
Step-by-step explanation:
Answer:
x=6
Step-by-step explanation:
Combine like terms, 5x and 7x
12x=72
Isolate the variable by dividing by 12
x=6
Answer:
ok
Step-by-step explanation:
ok ok ok ok ok ok ok ok
Answer:
(x -2) is a factor
Step-by-step explanation:
The factor theorem tells you that (x-2) is a factor of g(x) if g(2) = 0.
g(2) = 2·2³ -5·2² +2 +2 = 16 -20 +2 +2 = 0
Since g(2) = 0, (x -2) is a factor of g(x).
__
The graph shows us the complete factorization is ...
g(x) = (2x +1)(x -1)(x -2)
