Let h(x)=2^(x). Can we use the intermediate value theorem to say the equation h(x)=4 has a solution where 3

Question

Let  h(x)=2^(x). Can we use the intermediate value theorem to say the equation  h(x)=4 has a solution where  3 <= x <= 5 ? Choose 1 answer: (A) No, since the function is not continuous on that interval. (B) No, since 4 is not between  h(3) and  h(5). (C) Yes, both conditions for using the intermediate value theorem have been met.

Bytelearn · Accepted Answer

Recall IVT: First, let's recall the Intermediate Value Theorem (IVT). The IVT states that if a function f is continuous on a closed interval [a, b] and k is any number between f(a) and f(b), then there is at least one number c in the interval [a, b] such that f(c) = k. We need to check if the function h(x) = 2^x is continuous on the interval [3, 5] and if the value 4 lies between h(3) and h(5).Evaluate Endpoints: Now, let's evaluate the function h(x) at the endpoints of the interval. First, we calculate h(3) = 2^3. h(3) = 2^3 = 8Check Value 4: Next, we calculate h(5) = 2^5. h(5) = 2^5 = 32Continuity of h(x): We have h(3) = 8 and h(5) = 32. Now we need to check if the value 4 is between h(3) and h(5). Since 4 is less than 8 and 8 is less than 32, it is clear that 4 is indeed between h(3) and h(5).Apply IVT: The function h(x) = 2^x is an exponential function, which is known to be continuous everywhere on its domain, which includes all real numbers. Therefore, h(x) is continuous on the interval [3, 5].Since h(x) is continuous on [3, 5] and 4 is between h(3) and h(5), both conditions for using the Intermediate Value Theorem have been met. Therefore, we can conclude that there is at least one solution to the equation h(x) = 4 on the interval [3, 5].

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