1.Use the IVT to explain why the following is true. If f continuous on (a,b) and f(x) does not equal to zero for all x in the interval (a,b), then f(x) > for all x in the interval (a,b) or f(x) < 0 for all x in the interval (a,b).
2. You are given that a function f is defined and continuous at every real number x except x=2. Also, f(x)=0 precisely for x=1 and 7. Finally, you know that f(-3)=4, f(1.5)=2, f(4)= -3, and f(10)=6. Solve the inequality f(x) > 0.
1. To use the Intermediate Value Theorem (IVT) to explain why the given statement is true, let's analyze the conditions of the theorem.
The Intermediate Value Theorem states that if a function f(x) is continuous on a closed interval [a, b] and M is any value between f(a) and f(b), inclusive, then there exists at least one value c in the interval (a, b) such that f(c) = M.
In our case, we are given that f is continuous on the interval (a, b) and f(x) does not equal zero for all x in (a, b). We want to prove that either f(x) > 0 for all x in (a, b) or f(x) < 0 for all x in (a, b).
Assume, for contradiction, that there exists at least one value c in (a, b) such that f(c) > 0 and another value d in (a, b) such that f(d) < 0. Since f is continuous on (a, b), by the IVT, there must exist a value e in (a, b) such that f(e) = 0. However, this contradicts the given condition that f(x) does not equal zero for all x in (a, b). Hence, our assumption is false.
Therefore, either f(x) > 0 for all x in (a, b) or f(x) < 0 for all x in (a, b) must be true.
2. To solve the inequality f(x) > 0, we can use the given information about the function f. Let's analyze the problem step by step:
First, we know that f(x) = 0 precisely for x = 1 and 7. Therefore, these two points divide the real number line into three intervals: (-∞, 1), (1, 7), and (7, +∞).
Next, we need to determine the sign of f(x) in each interval. From the information given, we know that f(-3) = 4, f(1.5) = 2, f(4) = -3, and f(10) = 6.
In the interval (-∞, 1), f(x) is positive (greater than zero) because f(-3) = 4, which is greater than zero.
In the interval (1, 7), f(x) is negative (less than zero) because of the point x = 4, where f(4) = -3, which is less than zero.
In the interval (7, +∞), f(x) is positive (greater than zero) because f(10) = 6, which is greater than zero.
Therefore, the solution to the inequality f(x) > 0 is x ∈ (-∞, 1) U (7, +∞), which means x can be any real number except 1 and 7.
1. To use the Intermediate Value Theorem (IVT) to explain why the given statement is true, we assume that f is continuous on the interval (a, b) and f(x) does not equal zero for all x in the interval (a, b).
According to the IVT, if a function is continuous on a closed interval [a, b], and it takes on two different values, c and d, at the endpoints of the interval, then for any value between c and d, there exists at least one value within the interval where the function takes on that value.
In our case, since f(x) does not equal zero for all x in the interval (a, b), we can consider two cases:
Case 1: f(x) > 0 for some x in (a, b)
If there exists an x in (a, b) such that f(x) > 0, it means that the function f(x) takes on a positive value. Since f(x) is continuous, by the IVT, there must be some x within (a, b) where f(x) = 0, which contradicts the given condition. Therefore, this case is not possible.
Case 2: f(x) < 0 for all x in (a, b)
If f(x) < 0 for all x in (a, b), it means that the function f(x) takes on negative values throughout the interval (a, b). Since f(x) is continuous, by the IVT, there cannot be any x within (a, b) where f(x) = 0. In this case, the given statement holds true.
Therefore, based on the Intermediate Value Theorem, if f is continuous on (a, b) and f(x) does not equal zero for all x in (a, b), then f(x) must either be greater than zero for all x in (a, b) or less than zero for all x in (a, b).
2. To solve the inequality f(x) > 0, we need to find the set of values of x for which f(x) is greater than zero.
Given that f(x) = 0 for x = 1 and 7, we know that f(x) can only be positive or negative in the intervals where x is not equal to 1 or 7.
From the information given, we have the following points:
f(-3) = 4
f(1.5) = 2
f(4) = -3
f(10) = 6
Since f(-3) = 4 is positive and f(1.5) = 2 is positive, we can conclude that f(x) is positive in the interval (-3, 1.5).
Since f(4) = -3 is negative and f(10) = 6 is positive, we can conclude that f(x) is positive in the interval (4, 10).
Therefore, the solution to f(x) > 0 is the union of the intervals (-3, 1.5) and (4, 10), which can be written as (-3, 1.5) U (4, 10).