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.

__

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:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

after 1 year: 221 deer

step-by-step explanation:

30% of 170 is 51. 170+51=221