Functions
Piecewise Functions
GRB_1000_SCQ
Grade Class 11
Question:
If $f: R \to R$ is defined as $f(x) = \begin{cases} x+4, & x < -4 \\ 3x+2, & -4 \leq x < 4 \\ x-4, & x \geq 4 \end{cases}$, then the value of $f(f(f(f(0)))) + 1$ is equal to:
Step-by-Step Solution
Key Concept: Iterative evaluation of piecewise functions.
Step 1: Evaluate $f(0)$
We need to determine which piece of the piecewise function applies when $x = 0$.
Since $0 \in [-4, 4)$, we use the second piece: $f(x) = 3x + 2$
$$f(0) = 3(0) + 2 = 2$$
Step 2: Evaluate $f(f(0)) = f(2)$
Now we need to find $f(2)$.
Since $2 \in [-4, 4)$, we use the second piece: $f(x) = 3x + 2$
$$f(2) = 3(2) + 2 = 8$$
Step 3: Evaluate $f(f(f(0))) = f(8)$
Now we need to find $f(8)$.
Since $8 \geq 4$, we use the third piece: $f(x) = x - 4$
$$f(8) = 8 - 4 = 4$$
Step 4: Evaluate $f(f(f(f(0)))) = f(4)$
Now we need to find $f(4)$.
Since $4 \geq 4$, we use the third piece: $f(x) = x - 4$
$$f(4) = 4 - 4 = 0$$
Step 5: Calculate the final answer
We need to find $f(f(f(f(0)))) + 1$:
$$f(f(f(f(0)))) + 1 = 0 + 1 = 1$$
**Final Answer:** The value of $f(f(f(f(0)))) + 1$ is equal to $\boxed{1}$, which corresponds to **Option 2**.
Correct Answer: 2