Let h(x)=log_(2)(x)*x^(2). Below is Jasmine's attempt to write a formal justification for the fact that the equation h(x)=3 has a solution where 1

Question

Let  h(x)=log_(2)(x)*x^(2). Below is Jasmine's attempt to write a formal justification for the fact that the equation  h(x)=3 has a solution where  1 <= x <= 2. Is Jasmine's justification complete? If not, why? Jasmine's justification:  h(1)=0 and  h(2)=4, so 3 is between  h(1) and  h(2). So, according to the intermediate value theorem,  h(x)=3 must have a solution for an  x-value within the interval  [1,2]. Choose 1 answer: (A) Yes, Jasmine's justification is complete. (B) No, Jasmine didn't establish that 3 is between  h(1) and  h(2). (C) No, Jasmine didn't establish that  h is continuous.

Bytelearn · Accepted Answer

Understand Function and IVT: Understand the function and the Intermediate Value Theorem (IVT). The function h(x) = log_2(x) * x^2 is a product of a logarithmic function and a power function. The IVT states that if a function 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.Check Endpoint Values: Check the values of h(x) at the endpoints of the interval. Jasmine calculated h(1) and h(2). We need to verify these calculations. h(1) = log_2(1) * 1^2 = 0 * 1 = 0 h(2) = log_2(2) * 2^2 = 1 * 4 = 4 These calculations are correct, and they show that h(1) = 0 and h(2) = 4.Check Value 3: Determine if the value 3 is between h(1) and h(2). Since h(1) = 0 and h(2) = 4, the value 3 is indeed between h(1) and h(2). This part of Jasmine's justification is correct.Verify Continuity: Verify the continuity of the function h(x) on the interval [1, 2]. Jasmine's justification assumes that h(x) is continuous on the interval [1, 2] but does not explicitly state it. The function h(x) is continuous on the interval [1, 2] because both the logarithmic function log_2(x) and the power function x^2 are continuous on that interval, and the product of continuous functions is also continuous. However, Jasmine did not explicitly mention this in her justification.

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