Maple中函数逆的表达式序列处理：求导与分段函数转换问题

First, let's break down why you're seeing an expression sequence instead of a usable function: solving the quartic equation from $y = T(x)$ gives 4 algebraic roots, but only one valid real root lies within the domain $x \in [0,1]$ (since $T(x)$ is strictly increasing on $[0,1]$—we can confirm this by computing its derivative, which stays positive for all $x \in (0,1)$).

Here's a step-by-step fix to turn those roots into a differentiable/integratable function:

Step 1: Extract and Filter Valid Solutions #

First, capture all solutions in a list, then filter out those that don't satisfy $0 \leq x \leq 1$ (the domain of your original function):

# Define your original function
T := x-> sqrt(x)/(sqrt(x)+sqrt(1-x))^2;

# Get all solutions as a list, with assumptions on t
all_solutions := [solve(t = T(x), x, useassumptions=true) assuming 0<=t<=1];

# Filter to keep only valid x values in [0,1]
valid_solution := select(s -> evalb(0 <= s <= 1) assuming 0<=t<=1, all_solutions);

Since $T(x)$ is strictly increasing, valid_solution will be a list with exactly one element.

Step 2: Define a Usable Inverse Function #

Convert the valid solution into a proper function that Maple can work with for calculus operations:

# Define V(t) using the valid root
V := t -> eval(valid_solution[1], t = t);

# Test it (matches your observations)
V(0); # Returns 0
V(1); # Returns 1

Step 3: Compute Derivatives and Integrals #

Now you can use standard Maple calculus commands on $V(t)$ without errors:

# Compute derivative using diff
V_prime := diff(V(t), t);
# Or use the D operator for functional derivative
V_prime_op := D(V);

# Compute integral (e.g., definite integral from 0 to t)
V_integral := int(V(s), s=0..t);

What If There Were Multiple Valid Solutions? #

If your function weren't strictly monotonic (leading to multiple $x$ values for a single $t$), you'd convert it to a piecewise function:

1. Find critical $t$-values where the number of solutions changes (like the maximum/minimum of $T(x)$).
2. Define a piecewise function that selects the correct root for each interval:

   # Example structure (adjust based on your actual solutions)
   t_max := eval(T(x), x = solve(diff(T(x),x)=0, x));
   V_piecewise := piecewise(
       t <= t_max, solution1(t),
       t <= t_max, solution2(t),
       t > t_max, undefined
   );
   

But in your case, this isn't necessary since $T(x)$ is strictly increasing across its domain.

内容的提问来源于stack exchange，提问作者J.W