Question: A. Evaluate the Function
Given $f(x)=2x^2-3x+1$, find the following:
1. $f(2)$
2. $f(-1)$
3. $f(0)$
B. Determine the Domain
Find the domain of each function.
4. $f(x)=\frac{1}{x-5}$
5. $g(x)=2(x+8)$
6. $h(x)=\frac{3}{x^2-16}$
Graph shape: no visible graph; only a worksheet page with text and horizontal divider lines, position_hint=center
User: Answer correctly, please answer it right. Find the domain of each function
1. **Evaluate $f(2)$:**
Given $f(x)=2x^2-3x+1$, substitute $x=2$:
$$f(2)=2(2)^2 - 3(2) + 1 = 2 \times 4 - 6 + 1 = 8 - 6 + 1 = 3$$
2. **Evaluate $f(-1)$:**
Substitute $x=-1$:
$$f(-1)=2(-1)^2 - 3(-1) + 1 = 2 \times 1 + 3 + 1 = 2 + 3 + 1 = 6$$
3. **Evaluate $f(0)$:**
Substitute $x=0$:
$$f(0)=2(0)^2 - 3(0) + 1 = 0 - 0 + 1 = 1$$
---
4. **Find the domain of $f(x)=\frac{1}{x-5}$:**
- The denominator cannot be zero.
- Set denominator equal to zero and solve:
$$x - 5 = 0 \implies x = 5$$
- So, $x=5$ is excluded from the domain.
- **Domain:** All real numbers except $x=5$, or in interval notation:
$$(-\infty, 5) \cup (5, \infty)$$
5. **Find the domain of $g(x)=2(x+8)$:**
- This is a polynomial function (linear), which is defined for all real numbers.
- **Domain:** All real numbers, or:
$$(-\infty, \infty)$$
6. **Find the domain of $h(x)=\frac{3}{x^2-16}$:**
- Denominator cannot be zero.
- Set denominator equal to zero:
$$x^2 - 16 = 0$$
$$x^2 = 16$$
$$x = \pm 4$$
- Exclude $x=4$ and $x=-4$ from the domain.
- **Domain:** All real numbers except $x=4$ and $x=-4$, or:
$$(-\infty, -4) \cup (-4, 4) \cup (4, \infty)$$