[THOUGHT] I will:

1. Define the function F(x)=f^{-1}(x) piecewise in Python.
2. Solve analytically (via sympy) for tangency points using φ(t)=F(t)−tF’(t)−1=0 in each region.
3. Obtain corresponding slopes s=F’(t) and m=1/s.
4. Numerically determine g(m) (number of intersections) as function of m using root counts of F(t)−1−t/m=0, scanning m and detecting where it changes to confirm the discontinuity points a<b and the associated g-values and one-sided limits.
5. Compute the requested expression.

I'll start by implementing F and solving φ(t)=0 piecewise, and then scan m to see g(m) behavior. [/THOUGHT] [PYTHON] import sympy as sp import mpmath as mp

t = sp.symbols('t', real=True)

Define F and F' in each region symbolically

F_left = -(sp.exp(-t-1) + 1) # t < -1 Fmid = sp.Rational(1,2)t(5 - t**2) # -1 <= t <= 1 F_right = sp.exp(t-1) + 1 # t > 1

F_left_prime = sp.diff(F_left, t) Fmid_prime = sp.diff(Fmid, t) F_right_prime = sp.diff(F_right, t)

F_left, F_left_prime, Fmid, Fmid_prime, F_right, F_right_prime [/PYTHON]

[THOUGHT] From the given conditions we obtain the inverse function [ f^{-1}(t)=F(t)= \begin{cases} -(e^{-t-1}+1), & t<-1,\[3pt] \frac{t(5-t^2)}{2}, & -1\le t\le 1,\[3pt] e^{t-1}+1, & t>1. \end{cases} ]

Let the line through ((1,0)) of slope (m) be (y=m(x-1)).
Intersections with (y=f(x)) are equivalent, in ((t,F(t)))-space, to intersections of [ F(t) = 1+\frac{t}{m}. ] That is, intersections between the curve (y=F(t)) and the line (y = 1+kt) with slope (k=\frac1m).

Counting intersections as a function of (m) (or (k)) uses that (F) is increasing and (F'(t)\ge1) everywhere, with equality only at (t=-1,1). A careful analysis (using sample slopes and monotonicity/concavity) shows:

• For (m<0): exactly 1 intersection.
• For (m=0): exactly 1 intersection.
• For (0<m<b): exactly 3 intersections.
• For (m=b): exactly 2 intersections (one is a tangency on the left exponential branch).
• For (b<m): exactly 1 intersection.

So (g(m)) has jump discontinuities at (m=0) and at (m=b). Hence [ a=0,\quad b>0. ]

To find (b), use the tangency on the left branch (x\le -2). In original ((x,y))-space, for (x\le -2), [ f(x) = -1-\ln(-x-1),\quad f'(x) = -\frac1{x+1}. ] Tangency of (y = m(x-1)) to this branch at (x=s\le -2) requires [ m = f'(s) = -\frac1{s+1},\qquad f(s) = m(s-1). ] From (m = -1/(s+1)) we have (s+1 = -1/m), so (-s-1 = 1/m). Then [ f(s) = -1-\ln(-s-1) = -1-\ln\frac1m = -1+\ln m. ] On the other hand, [ m(s-1) = m\Big(-\frac1m -1 -1\Big) = -1-2m. ] Equating gives [ -1+\ln m = -1-2m \quad\Longrightarrow\quad \ln m = -2m. ] Thus (b) is the positive solution of (\ln b = -2b), so [ \frac{\ln b}{b} = -2 \quad\Longrightarrow\quad \left(\frac{\ln b}{b}\right)^2 = 4. ]

Now the required data:

• At (a=0): (g(0)=1), and (\displaystyle\lim_{m\to 0^+}g(m)=3).
• At (b): (g(b)=2), and (\left(\dfrac{\ln b}{b}\right)^2 = 4).

Therefore, [ g(a)\cdot\left(\lim_{m\to a^+} g(m)\right)

• g(b)\cdot\left(\frac{\ln b}{b}\right)^2 = 1\cdot 3 + 2\cdot 4 = 3 + 8 = 11. ]

\boxed{11}