Match the sequence (term) with the correct type of sequence (definition). 2, 4, 6, 8, ... Answer 1 28, 24, 20, 16, ... Answer 2
1 answer:
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Answer:
The equation does not have a real root in the interval
Step-by-step explanation:
We can make use of the intermediate value theorem.
The theorem states that if is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:
- If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
- The image of a continuous function over an interval is itself an interval.
Of course, in our case, we will make use of the first one.
First, we need to proof that our function is continues in , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval
, which means to evaluate the equation in 0 and 1:
Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval . I attached a plot of the equation in the interval
where you can clearly observe how the graph does not cross the x-axis in the interval.
