If f(x) = {3x−14x+2if −10<x<1if 2<x<10, then find the value of f(x) at x = -3.-8-10-12-14

The given function \( f(x) \) is written in a piecewise form, although it is currently not completely clear. To properly evaluate \( f(x) \) at the specified points, we need to understand the piecewise intervals and conditions for \( f(x) \). I will assume the function should be expressed like this:

\[
f(x) =
\begin{cases} 
3x - 14 & \text{if } -10 < x < 1, \\
x + 2 & \text{if } 2 < x < 10
\end{cases}
\]

Now, let's evaluate \( f(x) \) at each given \( x \)-value:

1. **At \( x = -3 \):**
   \[
   -10 < -3 < 1, \text{ so } f(x) = 3x - 14.
   \]
   \[
   f(-3) = 3(-3) - 14 = -9 - 14 = -23.
   \]

2. **At \( x = -8 \):**
   \[
   -10 < -8 < 1, \text{ so } f(x) = 3x - 14.
   \]
   \[
   f(-8) = 3(-8) - 14 = -24 - 14 = -38.
   \]

3. **At \( x = -10 \):**
   \( x = -10 \) is **not included** in the interval \( -10 < x < 1 \), so \( f(-10) \) is undefined.

4. **At \( x = -12 \):**
   \( x = -12 \) is **outside** both intervals \( -10 < x < 1 \) and \( 2 < x < 10 \), so \( f(-12) \) is undefined.

5. **At \( x = -14 \):**
   Similar to \( x = -12 \), \( x = -14 \) is also outside both intervals, so \( f(-14) \) is undefined.

### Summary:
- \( f(-3) = -23 \)
- \( f(-8) = -38 \)
- \( f(-10), f(-12), f(-14) \) are **undefined**.

Would you like more details, or do you have any questions?

Here are some related questions to explore:
1. What is the value of \( f(x) \) at \( x = 0 \)?
2. How do we interpret discontinuities in piecewise functions?
3. How would we rewrite \( f(x) \) if the intervals changed?
4. Can piecewise functions have overlapping intervals?
5. How do we handle boundary points in piecewise functions?

**Tip:** Always pay attention to whether intervals are open ( \( < \), \( > \) ) or closed ( \( \leq \), \( \geq \) ) to avoid including undefined points.

If f(x) = {3x−14 if −10 < x < 1, x+2 if 2 < x < 10}, then find the value of f(x) at x = -3, -8, -10, -12, -14.

Explore how to evaluate a piecewise function f(x) at specific x-values, including -3, -8, -10, -12, and -14. This problem highlights key algebraic concepts and the importance of understanding intervals. Suitable for students in Grades 9-11.

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