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].

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3

answer: ♡♡♡♡♡♡♡♡♡♡

step-by-step explanation: