Piecewise Function Worksheet - Fill and Sign Printable Template Online - Free Printable

Let’s solve each piecewise function step by step, determining whether it is a function, its domain and range, and evaluating specific values. We’ll also note if it’s continuous.

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Problem 1:

> \( f(x) = \begin{cases}
x - 8 & x < -2 \\
x^2 - 2x - 3 & x \geq -2
\end{cases} \)

#### Step 1: Is it a function?
Yes — for every input \(x\), there is exactly one output. The pieces are defined on non-overlapping domains (except at the boundary \(x = -2\), which is assigned to the second piece). So no ambiguity → Function: Yes

#### Step 2: Domain
All real numbers — the first piece covers \(x < -2\), the second covers \(x \geq -2\). Together, they cover all real numbers.

→ Domain: \( (-\infty, \infty) \)

#### Step 3: Range
We need to find the range of each piece.

- For \(x < -2\): \(f(x) = x - 8\) → linear, decreasing.
As \(x \to -\infty\), \(f(x) \to -\infty\)
At \(x = -2\) (approaching from left): \(f(-2^-) = -2 - 8 = -10\)
So this piece gives: \( (-\infty, -10) \)

- For \(x \geq -2\): \(f(x) = x^2 - 2x - 3 = (x-1)^2 - 4\)
This is a parabola opening upwards, vertex at \(x=1\), \(f(1) = -4\)
At \(x = -2\): \(f(-2) = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5\)
As \(x \to \infty\), \(f(x) \to \infty\)

So this piece gives: \([-4, \infty)\) — because minimum value is -4 at x=1, and at x=-2 it's 5, which is higher.

Now combine both ranges:
- First piece: \( (-\infty, -10) \)
- Second piece: \([-4, \infty)\)

There’s a gap between -10 and -4, so overall range is:

→ Range: \( (-\infty, -10) \cup [-4, \infty) \)

#### Step 4: Continuity
Check continuity at the boundary \(x = -2\):

- Left limit: \(\lim_{x \to -2^-} f(x) = -2 - 8 = -10\)
- Right limit: \(\lim_{x \to -2^+} f(x) = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5\)
- \(f(-2) = 5\) (since defined in second piece)

Left limit ≠ right limit → Not continuous at x = -2

→ Continuous: No

#### Step 5: Evaluate specific values

- \(f(-2)\): Use second piece → \( (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = \boxed{5} \)

- \(f(-4)\): Use first piece → \( -4 - 8 = \boxed{-12} \)

- \(f(2)\): Use second piece → \( 2^2 - 2(2) - 3 = 4 - 4 - 3 = \boxed{-3} \)

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Problem 2:

> \( f(x) = \begin{cases}
2x - 1 & x < 1 \\
x^2 - 3 & x \geq 1
\end{cases} \)

#### Step 1: Is it a function?
Yes — no overlapping definitions; at x=1, only second piece applies.

→ Function: Yes

#### Step 2: Domain
All real numbers — covers \(x < 1\) and \(x \geq 1\).

→ Domain: \( (-\infty, \infty) \)

#### Step 3: Range

- For \(x < 1\): \(f(x) = 2x - 1\) → linear, increasing.
As \(x \to -\infty\), \(f(x) \to -\infty\)
At \(x = 1^-\): \(f(1^-) = 2(1) - 1 = 1\)
So this piece gives: \( (-\infty, 1) \)

- For \(x \geq 1\): \(f(x) = x^2 - 3\) → parabola opening up, vertex at x=0 (but we start at x=1).
At x=1: \(f(1) = 1 - 3 = -2\)
As x increases, f(x) → ∞
Minimum in this interval is at x=1: -2

So this piece gives: \([-2, \infty)\)

Combine ranges:
- First piece: \( (-\infty, 1) \)
- Second piece: \([-2, \infty)\)

Overlap from -2 to 1 → combined range is:

→ Range: \( (-\infty, \infty) \) — since first piece goes to -∞, second to ∞, and they overlap.

Actually, let’s check: does it cover everything?

- From -∞ to 1: covered by first piece
- From -2 to ∞: covered by second piece
→ So yes, entire real line is covered.

→ Range: \( (-\infty, \infty) \)

#### Step 4: Continuity at x=1

- Left limit: \(\lim_{x \to 1^-} f(x) = 2(1) - 1 = 1\)
- Right limit: \(\lim_{x \to 1^+} f(x) = 1^2 - 3 = -2\)
- \(f(1) = -2\)

Left limit ≠ right limit → Not continuous

→ Continuous: No

#### Step 5: Evaluate values

- \(f(-2)\): Use first piece → \(2(-2) - 1 = -4 -1 = \boxed{-5}\)

- \(f(0)\): Use first piece → \(2(0) - 1 = \boxed{-1}\)

- \(f(2)\): Use second piece → \(2^2 - 3 = 4 - 3 = \boxed{1}\)

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Problem 3:

> \( f(x) = \begin{cases}
-2x + 1 & x < 2 \\
5x - 8 & x \geq 2
\end{cases} \)

#### Step 1: Is it a function?
Yes — no overlaps; x=2 is assigned to second piece.

→ Function: Yes

#### Step 2: Domain
All real numbers.

→ Domain: \( (-\infty, \infty) \)

#### Step 3: Range

- For \(x < 2\): \(f(x) = -2x + 1\) → linear, decreasing.
As \(x \to -\infty\), \(f(x) \to \infty\)
At \(x = 2^-\): \(f(2^-) = -2(2) + 1 = -4 + 1 = -3\)
So this piece gives: \( (-3, \infty) \)

- For \(x \geq 2\): \(f(x) = 5x - 8\) → linear, increasing.
At x=2: \(f(2) = 5(2) - 8 = 10 - 8 = 2\)
As \(x \to \infty\), \(f(x) \to \infty\)
So this piece gives: \([2, \infty)\)

Combine ranges:
- First piece: \( (-3, \infty) \)
- Second piece: \([2, \infty)\)

Combined: \( (-3, \infty) \) — since [2, ∞) is already included in (-3, ∞)

→ Range: \( (-3, \infty) \)

#### Step 4: Continuity at x=2

- Left limit: \(\lim_{x \to 2^-} f(x) = -2(2) + 1 = -3\)
- Right limit: \(\lim_{x \to 2^+} f(x) = 5(2) - 8 = 2\)
- \(f(2) = 2\)

Left limit ≠ right limit → Not continuous

→ Continuous: No

#### Step 5: Evaluate values

- \(f(-4)\): Use first piece → \(-2(-4) + 1 = 8 + 1 = \boxed{9}\)

- \(f(0)\): Use first piece → \(-2(0) + 1 = \boxed{1}\)

- \(f(2)\): Use second piece → \(5(2) - 8 = 10 - 8 = \boxed{2}\)

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## ✔ Final Answers Summary:

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Problem 1:

- Function: Yes
- Domain: \( (-\infty, \infty) \)
- Range: \( (-\infty, -10) \cup [-4, \infty) \)
- Continuous: No
- \(f(-2) = \boxed{5}\)
- \(f(-4) = \boxed{-12}\)
- \(f(2) = \boxed{-3}\)

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Problem 2:

- Function: Yes
- Domain: \( (-\infty, \infty) \)
- Range: \( (-\infty, \infty) \)
- Continuous: No
- \(f(-2) = \boxed{-5}\)
- \(f(0) = \boxed{-1}\)
- \(f(2) = \boxed{1}\)

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Problem 3:

- Function: Yes
- Domain: \( (-\infty, \infty) \)
- Range: \( (-3, \infty) \)
- Continuous: No
- \(f(-4) = \boxed{9}\)
- \(f(0) = \boxed{1}\)
- \(f(2) = \boxed{2}\)

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You can now sketch the graphs using these points and behaviors:

- Problem 1: Line with slope 1 going to (-2, -10), then parabola starting at (-2, 5) going down to vertex (1,-4) then up.
- Problem 2: Line with slope 2 going to (1,1), then parabola starting at (1,-2) going up.
- Problem 3: Line with slope -2 going to (2,-3), then line with slope 5 starting at (2,2).

All have jumps at the boundary point — hence discontinuous.

Let me know if you want graph sketches or further explanation!