To find the inverse function of $f(x) = \cos(x + 1)$, we need to follow these steps:

Step 1: Find the inverse function

1. Start with the function:
   $y = \cos(x + 1)$

2. Swap $x$ and $y$ to find the inverse:
   $x = \cos(y + 1)$

3. Solve for $y$:
   To isolate $y$, use the arccosine (inverse cosine) function:
   $y + 1 = \arccos(x)$
   $y = \arccos(x) - 1$

So, the inverse function is:
$f^{-1}(x) = \arccos(x) - 1$

Step 2: Determine the domain and range

• For the original function $f(x) = \cos(x + 1)$:

  • Domain: Since cosine is defined for all real numbers, the domain of $f(x)$ is $(-\infty, \infty)$.
  • Range: The cosine function outputs values between $-1$ and $1$, so the range of $f(x)$ is $[-1, 1]$.

• For the inverse function $f^{-1}(x) = \arccos(x) - 1$:

  • Domain: The arccosine function $\arccos(x)$ is defined for $x \in [-1, 1]$, so the domain of the inverse function is $[-1, 1]$.
  • Range: Since $\arccos(x)$ outputs values from $0$ to $\pi$, subtracting $1$ shifts this range to $[-1, \pi - 1]$.

Therefore:

• Inverse Function: $f^{-1}(x) = \arccos(x) - 1$
• Domain: $[-1, 1]$
• Range: $[-1, \pi - 1]$

Would you like more details or have any questions?

Here are some related questions:

1. What is the general method to find the inverse of a function?
2. How do transformations affect the domain and range of trigonometric functions?
3. What are the restrictions on the domain of inverse trigonometric functions?
4. How do you verify that two functions are inverses of each other?
5. Can the range of a function change if you modify its formula?

Tip: Remember, for inverse trigonometric functions, the range is adjusted based on principal values, so always consider these when determining inverses!