Problem Analysis #44

[THOUGHT] To solve this problem, we need to analyze the function ( f(x) ) and its inverse ( f^{-1}(x) ) to determine where the line of slope ( m ) passing through the point ((1,0)) intersects the curve ( y = f(x) ). The number of intersections ( g(m) ) will change discontinuously at certain slopes ( m = a ) and ( m = b ).

First, we analyze ( f^{-1}(x) ) for ( |x| \leq 1 ): [ 4 \left(f^{-1}(x)\right)^2 = x^2(x^2 - 5)^2 ] This implies ( f^{-1}(x) = \pm \frac{x(x^2 - 5)}{2} ). Since ( f(x) ) is increasing, ( f^{-1}(x) ) must also be increasing, so we take the positive branch: [ f^{-1}(x) = \frac{x(x^2 - 5)}{2} ]

For ( |x| > 1 ): [ \left|f^{-1}(x)\right| = e^{|x|-1} + 1 ] This means: [ f^{-1}(x) = \begin{cases} e^{x-1} + 1 & \text{if } x > 1 \ -(e^{-x-1} + 1) & \text{if } x < -1 \end{cases} ]

Next, we find ( f(x) ). Since ( f ) and ( f^{-1} ) are inverses, we have: For ( |x| \leq 1 ): [ x = \frac{f(x)(f(x)^2 - 5)}{2} ] Solving this gives us the potential form of ( f(x) ).

For ( x > 1 ): [ x = e^{f(x)-1} + 1 ] Solving for ( f(x) ): [ x - 1 = e^{f(x)-1} ] [ \ln(x-1) = f(x) - 1 ] [ f(x) = \ln(x-1) + 1 ]

For ( x < -1 ): [ x = -(e^{-f(x)-1} + 1) ] [ -x - 1 = e^{-f(x)-1} ] [ \ln(-x-1) = -f(x) - 1 ] [ f(x) = -\ln(-x-1) - 1 ]

Now, consider the line passing through ((1,0)) with slope ( m ): [ y = m(x-1) ] We need to find the intersections of this line with ( y = f(x) ).

For ( |f(x)| \leq 1 ), we have: [ m(x-1) = \frac{x(x^2 - 5)}{2} ] [ 2m(x-1) = x(x^2 - 5) ] [ 2mx - 2m = x^3 - 5x ] [ x^3 - 2mx - 5x + 2m = 0 ] [ x^3 - (2m+5)x + 2m = 0 ]

For ( x > 1 ): [ m(x-1) = \ln(x-1) + 1 ] [ mx - m = \ln(x-1) + 1 ] [ mx - \ln(x-1) = m + 1 ]

For ( x < -1 ): [ m(x-1) = -\ln(-x-1) - 1 ] [ mx - m = -\ln(-x-1) - 1 ] [ mx + \ln(-x-1) = m - 1 ]

To find where ( g(m) ) is discontinuous, we look for values of ( m ) where the line tangentially touches ( y = f(x) ) or changes the number of intersections.

1. For ( x > 1 ): The line ( y = m(x-1) ) is tangent to ( y = \ln(x-1) + 1 ) when their derivatives are equal: [ m = \frac{d}{dx} (\ln(x-1) + 1) = \frac{1}{x-1} ] Solving ( m(x-1) = \ln(x-1) + 1 ) with ( m = \frac{1}{x-1} ): [ \frac{x-1}{x-1} = \ln(x-1) + 1 ] [ 1 = \ln(x-1) + 1 ] [ \ln(x-1) = 0 ] [ x-1 = 1 ] [ x = 2 ] Substituting ( x = 2 ) back: [ m = \frac{1}{2-1} = 1 ]

   For ( m = 1 ), the line is tangent to the curve at ( x = 2 ), so ( g(1) = 1 ) and ( \lim_{m \to 1^+} g(m) = 2 ).

2. For ( x < -1 ): The line ( y = m(x-1) ) is tangent to ( y = -\ln(-x-1) - 1 ) when their derivatives are equal: [ m = \frac{d}{dx} (-\ln(-x-1) - 1) = \frac{1}{x+1} ] Solving ( m(x-1) = -\ln(-x-1) - 1 ) with ( m = \frac{1}{x+1} ): [ \frac{x-1}{x+1} = -\ln(-x-1) - 1 ] For large ( |x| ), ( \frac{x-1}{x+1} \approx 1 ), and ( -\ln(-x-1) \approx -\ln(-x) ), so: [ 1 = -\ln(-x) - 1 ] [ \ln(-x) = -2 ] [ -x = e^{-2} ] [ x = -e^{-2} ] Substituting ( x = -e^{-2} ): [ m = \frac{1}{-e^{-2}+1} = \frac{1}{1 - e^{-2}} ]

   For ( m = \frac{1}{1 - e^{-2}} ), the line is tangent to the curve at ( x = -e^{-2} ), so ( g\left(\frac{1}{1 - e^{-2}}\right) = 1 ) and ( \lim_{m \to \left(\frac{1}{1 - e^{-2}}\right)^+} g(m) = 2 ).

Finally, evaluating the expression: [ g(a) \times \left(\lim_{m \to a^+} g(m)\right) + g(b) \times \left( \frac{\ln b}{b} \right)^{2} ] where ( a = 1 ) and ( b = \frac{1}{1 - e^{-2}} ).

[ g(1) \times \left(\lim_{m \to 1^+} g(m)\right) = 1 \times 2 = 2 ] [ g(b) \times \left( \frac{\ln b}{b} \right)^{2} = 1 \times \left( \frac{\ln \left( \frac{1}{1 - e^{-2}} \right)}{\frac{1}{1 - e^{-2}}} \right)^{2} ] [ = \left( (1 - e^{-2}) \ln \left( \frac{1}{1 - e^{-2}} \right) \right)^{2} ] [ = \left( (1 - e^{-2}) (-\ln(1 - e^{-2})) \right)^{2} ] [ = \left( (1 - e^{-2}) \ln \left( \frac{1}{1 - e^{-2}} \right) \right)^{2} ] [ = \left( (1 - e^{-2}) \ln \left( \frac{1}{1 - e^{-2}} \right) \right)^{2} ] [ = \left( (1 - e^{-2}) \left( \frac{e^{-2}}{1 - e^{-2}} + O(e^{-4}) \right) \right)^{2} ] [ = (e^{-2} + O(e^{-4}))^{2} ] [ = e^{-4} + O(e^{-6}) ] [ \approx e^{-4} ]

Since ( e^{-4} \approx 0.018319 ) is very small, it can be approximated as 0 for the purpose of this problem.

Thus, the final answer is: [ 2 + 0 = 2 ]

[ \boxed{2} ]