The root x = 1 has multiplicity 3 for the function f(x)=x^3-x^2-x+1. True or false
1 answer:
X^3 - x^2 - x + 1 = 0
Plug in x = 1:-
1 -1 - 1 + 1 = 2 - 2 = 0 so x =1 is a root.
If it has multiplicity 3 then the function will factor to (x - 1)^3:-
(x - 1)^3 = (x - 1)(x^2 - 2x + 1) = x^3 - 2x^2 + x - x^2 + 2x - 1
= x^3 - 3x^2 + 3x - 1 so it is FALSE
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Answer:
100,000,000,009,099,998,878,374
Step-by-step explanation:
Area of a square = length²
A = x²
16 = x²; x = 4
dA/dx = 2x cm²/s
dx/dt = 6 cm/s
Using chain rule:
dA/dt = dA/dx * dx/dt
dA/dt = 2x * 6
dA/dt = 12x
At x = 4,
dA/dt = 12(4) = 48 cm²/s
This should be your answer
F(x) = 5(-3x) ^2 - (-3x) + 1
f(x) = -8x ^2 + 3x + 1
f(x) = -8x ^3 + 3x
