Question
\( \left. \begin{array} { l } { f ( x ) = \left\{ \begin{array} { l r } { 2 x ^ { 3 } - 1 , } & { x < 1 } \\ { 3 , } & { 1 \leq x < 5 } \\ { | x - 2 | , } & { x \geq 5 } \end{array} \right.} \\ { f ( 8 ) = } \\ { f ( 0 ) = } \\ { } \end{array} \right. \)
Ask by Powell Mathis. in the United States
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\( f(8) = 6 \) and \( f(0) = -1 \).
Solution
Let's evaluate the piecewise function \( f(x) \) at the specified points.
The function is defined as:
\[
f(x) =
\begin{cases}
2x^3 - 1 & \text{if } x < 1 \\
3 & \text{if } 1 \leq x < 5 \\
|x - 2| & \text{if } x \geq 5
\end{cases}
\]
### 1. Calculate \( f(8) \):
- **Determine the applicable case:**
Since \( 8 \geq 5 \), we use the third case: \( f(x) = |x - 2| \).
- **Compute \( f(8) \):**
\[
f(8) = |8 - 2| = |6| = 6
\]
**Thus, \( f(8) = 6 \).**
### 2. Calculate \( f(0) \):
- **Determine the applicable case:**
Since \( 0 < 1 \), we use the first case: \( f(x) = 2x^3 - 1 \).
- **Compute \( f(0) \):**
\[
f(0) = 2(0)^3 - 1 = 0 - 1 = -1
\]
**Thus, \( f(0) = -1 \).**
### Summary:
\[
\begin{align*}
f(8) &= 6 \\
f(0) &= -1
\end{align*}
\]
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Mind Expander
To find \( f(8) \), we look at the piece of the function that applies when \( x \geq 5 \). According to the given function \( f(x) = |x - 2| \) for \( x \geq 5 \). Thus, \[ f(8) = |8 - 2| = |6| = 6. \] Next, for \( f(0) \), we see that \( 0 < 1 \), which falls under the first case of the piecewise function. Thus, \[ f(0) = 2(0)^3 - 1 = 0 - 1 = -1. \] So, the final answers are: \( f(8) = 6 \) and \( f(0) = -1 \).
