[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

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

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

To find $b$, use the tangency on the left branch $x\le -2$. In original $(x,y)$-space, for $x\le -2$,

Tangency of $y = m(x-1)$ to this branch at $x=s\le -2$ requires

From $m = -1/(s+1)$ we have $s+1 = -1/m$, so $-s-1 = 1/m$. Then

On the other hand,

Equating gives

Thus $b$ is the positive solution of $\ln b = -2b$, so

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,

$\boxed{11}$