If f:[-6,6] rightarrow R is defined by f(x)=x | vedclass

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(A) Given, $f(x)=x^2-3$. First, we calculate $(f \circ f \circ f)(-1)$: $f(-1) = (-1)^2 - 3 = -2$ $f(f(-1)) = f(-2) = (-2)^2 - 3 = 1$ $f(f(f(-1))) = f

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$f(4 \sqrt{2})$

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(A) Given, $f(x)=x^2-3$. First, we calculate $(f \circ f \circ f)(-1)$: $f(-1) = (-1)^2 - 3 = -2$ $f(f(-1)) = f(-2) = (-2)^2 - 3 = 1$ $f(f(f(-1))) = f(1) = (1)^2 - 3 = -2$. Next, we calculate $(f \circ f \circ f)(0)$: $f(0) = (0)^2 - 3 = -3$ $f(f(0)) = f(-3) = (-3)^2 - 3 = 6$ $f(f(f(0))) = f(6) = (6)^2 - 3 = 33$. Next, we calculate $(f \circ f \circ f)(1)$: $f(1) = (1)^2 - 3 = -2$ $f(f(1)) = f(-2) = (-2)^2 - 3 = 1$ $f(f(f(1))) = f(1) = (1)^2 - 3 = -2$. Summing these values: $(f \circ f \circ f)(-1) + (f \circ f \circ f)(0) + (f \circ f \circ f)(1) = -2 + 33 - 2 = 29$. Now, checking the options: $f(4 \sqrt{2}) = (4 \sqrt{2})^2 - 3 = 32 - 3 = 29$. Thus, the expression is equal to $f(4 \sqrt{2})$.

If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then $(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)$ is equal to

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