$$ \left. \begin{array} { l } { f ( x ) = \left\{ \begin{array} { l r } { 2 x ^ { 3 } - 1 , } & { x

Question

$$ \left. \begin{array} { l } { f ( x ) = \left\{ \begin{array} { l r } { 2 x ^ { 3 } - 1 , } & { x < 1 } \ { 3 , } & { 1 \leq x < 5 } \ { | x - 2 | , } & { x \geq 5 } \end{array} ight.} \ { = - 1 \quad | f ( 4 ) = 5 } \end{array} ight. $$

UpStudy AI · Accepted Answer

Let's analyze the piecewise function you've provided and examine the conditions you've mentioned.

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

### Conditions:
1. $$ f(x) = -1 $$
2. $$ f(4) = 5 $$

Let's address each condition separately.

---

### 1. Solving $$ f(x) = -1 $$

We need to find all $$ x $$ such that $$ f(x) = -1 $$.

**Case 1: $$ x < 1 $$**
$$
f(x) = 2x^3 - 1 = -1 \
2x^3 = 0 \
x^3 = 0 \
x = 0
$$
**Solution:** $$ x = 0 $$

**Case 2: $$ 1 \leq x < 5 $$**
$$
f(x) = 3
$$
Since $$ 3 \neq -1 $$, there are no solutions in this interval.

**Case 3: $$ x \geq 5 $$**
$$
f(x) = |x - 2| = -1
$$
The absolute value $$ |x - 2| $$ is always non-negative, so it cannot equal $$-1$$.

**Conclusion for $$ f(x) = -1 $$:**
$$
\boxed{x = 0}
$$

---

### 2. Evaluating $$ f(4) = 5 $$

Let's compute $$ f(4) $$ using the given piecewise function.

**Determine the Interval for $$ x = 4 $$:**
$$
1 \leq 4 < 5
$$
So, we use:
$$
f(4) = 3
$$

**Given Condition:**
$$
f(4) = 5
$$

**Analysis:**
According to the piecewise definition, $$ f(4) = 3 $$. However, you've stated that $$ f(4) = 5 $$. This presents a contradiction because, within the interval $$ 1 \leq x < 5 $$, $$ f(x) $$ is consistently defined as $$ 3 $$.

**Possible Resolutions:**
1. **Typographical Error:** Ensure that the function definition or the condition $$ f(4) = 5 $$ is correctly stated. There might be a typo in either the function or the condition.
   
2. **Function Adjustment:** If $$ f(4) = 5 $$ is a necessary condition, you might need to redefine the function to accommodate this. For example, you could modify the piecewise definition to include an exception at $$ x = 4 $$:
   $$
   f(x) = \begin{cases} 
   2x^3 - 1, & \text{if } x < 1 \
   \begin{cases}
      5, & \text{if } x = 4 \
      3, & \text{otherwise}
   \end{cases}, & \text{if } 1 \leq x < 5 \
   |x - 2|, & \text{if } x \geq 5 
   \end{cases}
   $$
   
3. **Clarify Requirements:** If there's additional context or constraints, please provide them so that a more accurate solution can be formulated.

---

### Summary

- **Solution to $$ f(x) = -1 $$:** $$ x = 0 $$
- **Issue with $$ f(4) = 5 $$:** Contradicts the given function definition where $$ f(4) = 3 $$. Consider verifying the function or the condition.

If you intended something different or need further assistance with adjustments, please provide additional details!
$$ x = 0 $$ is the solution for $$ f(x) = -1 $$. However, $$ f(4) = 3 $$, not 5, based on the given function.

\( \left. \begin{array} { l } { f ( x ) = \left\{ \begin{array} { l r } { 2 x ^ { 3 } - 1 , } & { x < 1 } \\ { 3 , } & { 1 \leq x < 5 } \\ { | x - 2 | , } & { x \geq 5 } \end{array} \right.} \\ { = - 1 \quad | f ( 4 ) = 5 } \end{array} \right. \)

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