Question - Understanding Functions and Their Inverses
Solution:
The image displays two functions, h and g. The function h is defined as h: x → 2x+1, for real numbers x such that 0 ≤ x ≤ k, and the function g is defined as g: x → 2x for 0 ≤ x ≤ 1/2 and 2 - 2x for 1/2 ≤ x ≤ 1.Part i of the question asks: Given that the function h has an inverse, state the greatest value of k. Find h⁻¹(½) and g(h⁻¹(½)).To solve part i, let's break it down:1. **Function h having an inverse:** A function has an inverse if and only if it is one-to-one (injective), which means it never assigns the same value to two different domain elements. The function h given by h(x) = 2x + 1 is a linear function with a positive slope, which is injective over the entire domain of real numbers. Thus, there is no restriction on the value of k for h to have an inverse, and the greatest value of k is the largest value in the domain of x given for h, which is ∞. However, given the typical constraints of such problems, the greatest value of k could be considered the maximum value for which h is defined, which would be k if we associate it with the boundary of the domain. Since the problem does not specify an upper limit to the domain, we assume k can grow indefinitely.2. **Finding h⁻¹(½):** The inverse function, h⁻¹(x), will undo the operation performed by h(x). That is, if h(x) = 2x + 1, then for h⁻¹(y), where y = h(x), we must have x = (y - 1)/2. So, h⁻¹(½) means we should solve (½ - 1)/2 for x:h⁻¹(½) = (½ - 1)/2 = (-½)/2 = -¼3. **Finding g(h⁻¹(½)):** Now that we have x = -¼, we need to determine which piece of the piecewise function g to use. Since -¼ is not within the domain of either piece of g as defined, g(h⁻¹(½)) is not directly determined from the given definition of g and this seems to be an issue with either the domain of g as presented or the parameters of the problem. Typically, with the domain given for g, it's likely intended for positive values of x, and h⁻¹(½) gives us a negative value. Thus, there must be an assumption or additional information about function g beyond the given domain, or an error in the problem as written.In conclusion, for part i, the greatest value of k isn't constrained by the information given, and to find g(h⁻¹(½)), we would need further clarification on the behavior of g beyond the domain provided in the image.
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