这题怎么做？可以用计算器---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 have a function analysis problem involving exponential functions. Function f of x equals e to the x plus e to the negative x. Function h of x equals one over k times f of k minus x, where k is a positive parameter. We need to find the smallest value of k such that h intersects with the inverse of h sub one. The answer is one point two seven. Now let's understand function h of x. We can rewrite h of x as one over k times the sum of e to the k minus x plus e to the x minus k. This simplifies to two over k times hyperbolic cosine of x minus k. The function h sub one is the restriction of h to the domain from k to infinity. For x greater than or equal to k, the range of h sub one is from two over k to infinity. To find when h intersects with its inverse, we look for points where x equals h sub one of x. This means the graphs intersect on the line y equals x. Substituting u equals x minus k, we get k times k plus u equals two cosh u. We define F of u as two cosh u minus k u minus k squared. The minimum occurs when the derivative equals zero, giving us u sub k equals arc sinh of k over two. We need to find the smallest value of k such that the function h of x intersects with the inverse of h sub one. Let's start by understanding our given functions. To find where h of x intersects with h sub one inverse, we set them equal. This gives us a transcendental equation. For the minimum value of k, the intersection occurs when the functions are tangent to each other. We compute the derivatives of both functions. The derivative of h is straightforward. For the inverse function, we use the inverse function theorem: the derivative of the inverse equals one over the derivative of the original function. The tangency condition gives us a function F of k that must equal zero. Using numerical methods, we find that F of one point two six is positive while F of one point two seven is negative. Therefore, the critical value k sub zero is approximately one point two seven. To summarize: we found the intersection condition, applied the tangency requirement for the minimum k value, solved the resulting transcendental equation numerically, and determined that k sub zero equals approximately one point two seven. To summarize our solution: we found the intersection condition between h of x and h sub one inverse. We applied the tangency condition to find the minimum k value. We solved the resulting transcendental equation numerically and determined that the smallest value of k is one point two seven.