这题怎么做？---Let $f: R \rightarrow R, f(x) = e^x + e^{-x}$ and $g: R \rightarrow R, g(x) = \frac{1}{2}f(2-x)$. Let $h: R \rightarrow R, h(x) = \frac{1}{k}f(k-x)$, where $k \in (0, \infty)$. Let $h_1: [k, \infty) \rightarrow R, h_1(x) = h(x)$. The rule for the inverse of $h_1$ is $y = \log_{e} \left(\frac{k}{2}x + \frac{1}{2}\sqrt{k^2x^2 - 4}\right) + k$. e. What is the smallest value of $k$ such that $h$ will intersect with the inverse of $h_1$? Give your answer correct to two decimal places. [Horizontal line] [Horizontal line] [Horizontal line]

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We need to find the smallest value of k such that function h of x intersects with the inverse of h sub one. Given f of x equals e to the x plus e to the negative x, we define h of x as one over k times f of k minus x. The function h sub one is the restriction of h to the interval from k to infinity, and we're given the formula for its inverse. To find when the functions intersect, we first analyze the domain of h sub one inverse. The function h sub one is the restriction of h to the interval from k to infinity. The minimum value occurs at x equals k, giving us h sub one of k equals two over k. Since h sub one is increasing for x greater than or equal to k, the range is from two over k to infinity. Therefore, the domain of h sub one inverse is from two over k to infinity. Now we analyze the critical boundary condition. At x equals two over k, the inverse function h sub one inverse equals k. For the functions to intersect at this boundary, we need h of two over k to equal k. This leads to the equation e to the power of k minus two over k plus e to the power of two over k minus k equals k squared. Using the substitution a equals k minus two over k, we get e to the a plus e to the negative a equals k squared. Since the minimum value of e to the a plus e to the negative a is 2, occurring when a equals 0, we need k squared greater than or equal to 2, which gives us k greater than or equal to square root of 2. Let's verify our answer. When k equals square root of 2, which is approximately 1.414, we have a equals k minus two over k equals square root of 2 minus two over square root of 2, which equals zero. This gives us e to the zero plus e to the zero equals 2, which equals square root of 2 squared. The functions intersect at the point square root of 2, square root of 2. Therefore, the smallest value of k for which h of x intersects with the inverse of h sub one is square root of 2, which rounds to 1.41. To summarize our solution: We analyzed the intersection of function h with the inverse of h sub one by examining the boundary condition at x equals two over k. This led to the critical equation e to the a plus e to the negative a equals k squared. Using the fundamental property that the minimum value of this expression is 2, we determined that k must be at least square root of 2. Therefore, the smallest value of k for intersection is square root of 2, which equals 1.41 to two decimal places.