TRY IT! $$ \left.\begin{array}{lr}f(x)=\left\{\begin{array}{lr}2 x^{3}-1, & x

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TRY IT! $$ \left.\begin{array}{lr}f(x)=\left\{\begin{array}{lr}2 x^{3}-1, & x<1 \ 3, & 1 \leq x<5 \ |x-2|, & x \geq 5\end{array}\right. \ f(8)=60 & f(0)=-1\end{array} \right\rvert\, f(4)= $$

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To find $$ f(4) $$ using the piecewise function defined as:

$$
f(x) = 
\begin{cases} 
2x^3 - 1 & \text{if } x < 1 \ 
3 & \text{if } 1 \leq x < 5 \ 
|x - 2| & \text{if } x \geq 5 
\end{cases}
$$

we need to determine which condition $$ x = 4 $$ falls under.

1. **Identify the interval for $$ x = 4 $$**:
   - The first case $$ 2x^3 - 1 $$ applies for $$ x < 1 $$.
   - The second case $$ 3 $$ applies for $$ 1 \leq x < 5 $$.
   - The third case $$ |x - 2| $$ applies for $$ x \geq 5 $$.

Since $$ 4 $$ falls within the interval $$ 1 \leq x < 5 $$, we use the second case.

2. **Evaluate $$ f(4) $$**:
   - According to the second case, $$ f(4) = 3 $$.

Thus, the final answer is:

$$
f(4) = 3
$$
$$ f(4) = 3 $$

TRY IT! \[ \left.\begin{array}{lr}f(x)=\left\{\begin{array}{lr}2 x^{3}-1, & x<1 \\ 3, & 1 \leq x<5 \\ |x-2|, & x \geq 5\end{array}\right. \\ f(8)=60 & f(0)=-1\end{array} \right\rvert\, f(4)= \]

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