Prove that the equation
x^3 + x + 3 = 0
has exactly one real root on the interval [-2,-1].

Step-by-step explanation:

All polynomial functions are continuous, so we know from the intermediate value theorem that if the expression on the left changes sign in the interval [-2, 1] then there will be a zero in that interval. If the function is monotonic in the interval, there can only be one zero.

a) For f(x) = x^3 +x +3 = (x^2 +1)x +3, the values at the ends of the interval are ...

  f(-2) = (4+1)(-2) +3 = -7

  f(-1) = (1 +1)(-1) +3 = 1

The function value goes from -7 to +1 in the interval, so there exists at least one root in that interval.

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b) The derivative of the function is ...

  f'(x) = 3x^2 +1

This is positive for any x, so is positive in the interval [-2, -1]. That is, the function is continuously increasing in that interval, so cannot have more than one crossing of the x-axis. There is exactly one root in the interval [-2, -1].

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb