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How do you use the intermediate value theorem to show that there is a root of the equation 2x³ +x² +2=0 over the interval (−2,−1) ?

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User Gmo
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Final answer:

The intermediate value theorem states that if a continuous function takes on two distinct values at two points in an interval, then it must also take on every value between those two points in the interval. To show that there is a root of the equation 2x³ + x² + 2 = 0 over the interval (-2, -1), we need to show that the function f(x) = 2x³ + x² + 2 changes sign from negative to positive or vice versa within that interval.

Step-by-step explanation:

The intermediate value theorem states that if a continuous function takes on two distinct values at two points in an interval, then it must also take on every value between those two points in the interval. To show that there is a root of the equation 2x³ + x² + 2 = 0 over the interval (-2, -1), we need to show that the function f(x) = 2x³ + x² + 2 changes sign from negative to positive or vice versa within that interval.

First, calculate f(-2) and f(-1) to determine their signs. Substitute -2 and -1 into the equation:

f(-2) = 2(-2)³ + (-2)² + 2 = -12

f(-1) = 2(-1)³ + (-1)² + 2 = 1

Since f(-2) = -12 is negative and f(-1) = 1 is positive, the function must change sign somewhere between -2 and -1. Hence, there must be at least one root of the equation 2x³ + x² + 2 = 0 in the interval (-2, -1).

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User Miqwit
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