To solve the problem of finding the maximum and minimum values of the expression \( | \sin^{-1} x | + | \cos^{-1} x | \), we can follow these steps:
### Step 1: Understand the ranges of the functions
The functions \( \sin^{-1} x \) and \( \cos^{-1} x \) are defined for \( x \) in the interval \([-1, 1]\). Specifically:
- \( \sin^{-1} x \) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
- \( \cos^{-1} x \) ranges from \(0\) to \(\pi\).
### Step 2: Analyze the expression in different intervals
We will analyze the expression \( | \sin^{-1} x | + | \cos^{-1} x | \) in two intervals:
1. For \( x \in [0, 1] \)
2. For \( x \in [-1, 0] \)
#### Case 1: \( x \in [0, 1] \)
In this interval:
- \( \sin^{-1} x \) is non-negative, so \( | \sin^{-1} x | = \sin^{-1} x \).
- \( \cos^{-1} x \) is also non-negative, so \( | \cos^{-1} x | = \cos^{-1} x \).
Thus, the expression simplifies to:
\[
f(x) = \sin^{-1} x + \cos^{-1} x
\]
Using the identity \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \), we find:
\[
f(x) = \frac{\pi}{2}
\]
#### Case 2: \( x \in [-1, 0] \)
In this interval:
- \( \sin^{-1} x \) is negative, so \( | \sin^{-1} x | = -\sin^{-1} x \).
- \( \cos^{-1} x \) is non-negative, so \( | \cos^{-1} x | = \cos^{-1} x \).
Thus, the expression simplifies to:
\[
f(x) = -\sin^{-1} x + \cos^{-1} x
\]
Using the identity \( \cos^{-1} x = \pi - \sin^{-1}(-x) \), we can rewrite this as:
\[
f(x) = -\sin^{-1} x + \pi - \sin^{-1}(-x) = \pi - 2\sin^{-1}(-x)
\]
Since \( \sin^{-1}(-x) = -\sin^{-1}(x) \), we have:
\[
f(x) = \pi + 2\sin^{-1} x
\]
### Step 3: Find the maximum and minimum values
1. For \( x \in [0, 1] \), \( f(x) = \frac{\pi}{2} \).
2. For \( x \in [-1, 0] \):
- At \( x = -1 \), \( f(-1) = \pi + 2 \cdot (-\frac{\pi}{2}) = 0 \).
- At \( x = 0 \), \( f(0) = \pi + 2 \cdot 0 = \pi \).
Thus, the minimum value \( m = 0 \) and the maximum value \( M = \pi \).
### Step 4: Calculate \( M + m \)
Now, we can find:
\[
M + m = \pi + 0 = \pi
\]
### Final Answer
Therefore, the sum of the maximum and minimum values is:
\[
\boxed{\pi}
\]
