A continuous function h(x) has the values:
x h(x)
1 1
2 -4
3 5
4 -1
5 3
6 -2
7 -1
a. How many roots does h(x) have between x = 1 and x = 5?
b. How many roots does h(x) have between x = 1 and x = 7?
c. In how many places can h(x) = 3.4 between x = 1 and x = 7?
d. In how many places can h(x) = -0.5 between x = 1 and x = 7?
e. Is it possible that h(x) = π for some value(s) of x between 5 and 6?
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To find the number of roots of a continuous function, we need to look for sign changes in the function. Let's analyze each question:
a. Between x = 1 and x = 5, we have h(1) = 1 and h(5) = 3. Since there is a sign change, there must be at least one root in this interval.
b. Between x = 1 and x = 7, we have h(1) = 1 and h(7) = -1. Since there is a sign change, there must be at least one root in this interval.
c. To find the number of places where h(x) = 3.4 between x = 1 and x = 7, we need to look for sign changes around 3.4. From the given values, we can see that h(3) = 5 is greater than 3.4, while h(4) = -1 is less than 3.4. Therefore, there is exactly one place where h(x) = 3.4 in this interval.
d. Similar to part c, to find the number of places where h(x) = -0.5 between x = 1 and x = 7, we need to look for sign changes around -0.5. From the given values, we can see that h(2) = -4 is less than -0.5, while h(3) = 5 is greater than -0.5. Therefore, there is exactly one place where h(x) = -0.5 in this interval.
e. No, it is not possible for h(x) to equal π because the given values for h(x) are all integers. Therefore, h(x) cannot be equal to a non-integer value like π.
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