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:
0
1
2
4

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

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