Okay, so I need to find the range of the function f(x) = (arccos x)^2 + (arcsin x)^2. Hmm, let's start by recalling some properties of the arccos and arcsin functions. 

First, I know that arccos x and arcsin x are inverse functions, and their domains are both [-1, 1]. Also, there's a relationship between them: arccos x + arcsin x = π/2. Wait, is that right? Let me double-check. If x is in [-1, 1], then arccos x gives an angle between 0 and π, and arcsin x gives an angle between -π/2 and π/2. But when you add them together, does it equal π/2? Let me test with a value. Let's take x = 0. arccos 0 is π/2, and arcsin 0 is 0. So π/2 + 0 = π/2, which is indeed π/2. Okay, so that relationship holds. So arccos x + arcsin x = π/2. That's a key identity here.

So, given that, maybe I can express one in terms of the other. Let me denote θ = arccos x. Then, arcsin x = π/2 - θ. So substituting into the function f(x), we get f(x) = θ² + (π/2 - θ)². Let's expand that.

Expanding (π/2 - θ)² gives (π/2)^2 - πθ + θ². So adding θ² to that, we have θ² + (π/2)^2 - πθ + θ² = 2θ² - πθ + (π/2)^2. So f(x) simplifies to 2θ² - πθ + (π/2)^2 where θ is arccos x. But since θ = arccos x, and arccos x ranges from 0 to π as x ranges from 1 to -1, θ ∈ [0, π]. So θ is a variable in [0, π], and we need to find the range of the quadratic function 2θ² - πθ + (π/2)^2 over that interval.

Wait, so if I consider θ as a variable from 0 to π, then f(x) is a quadratic in θ. Let me denote g(θ) = 2θ² - πθ + (π/2)^2. To find the range of g(θ) over θ ∈ [0, π], we can analyze this quadratic function. Since the coefficient of θ² is positive (2), the parabola opens upwards, so the minimum occurs at the vertex, and the maximum occurs at one of the endpoints.

First, let's find the vertex of the parabola. The vertex occurs at θ = -b/(2a) = π/(2*2) = π/4. So the vertex is at θ = π/4. Let's compute g(π/4):

g(π/4) = 2*(π/4)^2 - π*(π/4) + (π/2)^2
= 2*(π²/16) - π²/4 + π²/4
= (π²/8) - π²/4 + π²/4
= π²/8

So the minimum value of g(θ) is π²/8. Now, let's check the endpoints θ = 0 and θ = π.

At θ = 0:
g(0) = 2*0 - π*0 + (π/2)^2 = π²/4

At θ = π:
g(π) = 2*(π)^2 - π*(π) + (π/2)^2
= 2π² - π² + π²/4
= π² + π²/4
= 5π²/4

So the maximum value of g(θ) is 5π²/4. Therefore, the range of f(x) is [π²/8, 5π²/4].

Wait, but let me double-check these calculations. Let's verify each step.

First, expanding (π/2 - θ)^2:
(π/2 - θ)^2 = (π/2)^2 - 2*(π/2)*θ + θ² = π²/4 - πθ + θ². Then, adding θ² gives θ² + (π²/4 - πθ + θ²) = 2θ² - πθ + π²/4. That seems correct.

Then, the quadratic in θ is 2θ² - πθ + π²/4. The vertex is at θ = π/(4). Plugging back in:

2*(π/4)^2 = 2*(π²/16) = π²/8
- π*(π/4) = -π²/4
+ π²/4. So total is π²/8 - π²/4 + π²/4 = π²/8. Correct.

At θ = 0: 0 + 0 + π²/4 = π²/4. Correct.

At θ = π: 2π² - π² + π²/4 = (2π² - π²) + π²/4 = π² + π²/4 = 5π²/4. Correct.

So the range is from π²/8 to 5π²/4. Therefore, the range of f(x) is [π²/8, 5π²/4].

Wait, but is there any possibility that this is incorrect? Let me check with specific values of x.

For example, when x = 1: arccos 1 = 0, arcsin 1 = π/2. So f(1) = 0 + (π/2)^2 = π²/4. Which matches the endpoint at θ = 0. Similarly, when x = -1: arccos(-1) = π, arcsin(-1) = -π/2. But arcsin(-1) is -π/2, but when we square it, it becomes (π/2)^2. Wait, hold on. Wait, arcsin(-1) is -π/2, but in the original function, it's (arcsin x)^2. So regardless of the sign, it's squared. So for x = -1, arcsin(-1)^2 = (-π/2)^2 = π²/4. Similarly, arccos(-1) = π, so (arccos(-1))^2 = π². So f(-1) = π² + π²/4 = 5π²/4. That's the other endpoint. So that's consistent.

Another test: when x = 0, arccos 0 = π/2, arcsin 0 = 0. So f(0) = (π/2)^2 + 0 = π²/4. Wait, but according to our earlier analysis, when θ = π/2, which is in the middle of the interval [0, π], the value is π²/8. But when x = 0, θ = π/2, which gives f(x) = π²/4. Wait, that's a discrepancy. Wait, hold on. Wait, no. Wait, when θ = π/2, which is arccos x = π/2, then x = cos(π/2) = 0. So f(x) = (arccos 0)^2 + (arcsin 0)^2 = (π/2)^2 + 0 = π²/4. But according to our earlier substitution, when θ = π/2, g(θ) = 2*(π/2)^2 - π*(π/2) + (π/2)^2.

Let's compute that: 2*(π²/4) - π²/2 + π²/4 = π²/2 - π²/2 + π²/4 = π²/4. Which matches. So the value at θ = π/2 is π²/4, which is the same as the value at θ = 0. Wait, but according to our quadratic function, the minimum is at θ = π/4, which gives π²/8. So when θ = π/4, which is arccos x = π/4, then x = cos(π/4) = √2/2. Then f(x) = (π/4)^2 + (π/2 - π/4)^2 = (π²/16) + (π/4)^2 = π²/16 + π²/16 = π²/8. Which is correct. So that's the minimum.

So, when θ = π/2, which is at x = 0, the value is π²/4, which is higher than π²/8. So the function f(x) reaches its minimum at θ = π/4 (x = √2/2) and maximum at θ = π (x = -1). The value at θ = 0 (x = 1) is π²/4, which is between the minimum and maximum. So the range is indeed from π²/8 to 5π²/4.

Therefore, the range of f(x) is [π²/8, 5π²/4]. So in boxed form, that would be [\frac{\pi^2}{8}, \frac{5\pi^2}{4}].

But just to make sure I didn't make any mistakes in the algebra. Let me recheck the quadratic function:

We had f(x) = (arccos x)^2 + (arcsin x)^2. Let θ = arccos x. Then arcsin x = π/2 - θ. So f(x) = θ² + (π/2 - θ)^2. Expanding that:

θ² + (π²/4 - πθ + θ²) = 2θ² - πθ + π²/4. Correct.

Then, treating this as a quadratic in θ: 2θ² - πθ + π²/4. The vertex is at θ = π/(4), which is correct. Then plugging in θ = π/4 gives 2*(π²/16) - π*(π/4) + π²/4 = π²/8 - π²/4 + π²/4 = π²/8. Correct.

Endpoints at θ = 0: 0 + 0 + π²/4 = π²/4. Correct.

At θ = π: 2π² - π² + π²/4 = π² + π²/4 = 5π²/4. Correct.

Therefore, the range is [π²/8, 5π²/4]. Seems solid. So I think that's the answer.

**Final Answer**
The range of \( f(x) \) is \boxed{\left[ \frac{\pi^2}{8}, \frac{5\pi^2}{4} \right]}.
Therefore, the range of $f(x)$ is $\boxed{\left[ \frac{\pi^2}{8}, \frac{5\pi^2}{4} \right]}$.
</think>