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Quiz 5
1. Define $f(x)=2x^2+3x+1$. Calculate $f(3)$.
Solution: Calculate
$$f(3)=2(3)^2+3(3)+1=18+9+1=28.$$
2. Define $f(x)=2x^2+3x+1$. Calculate $f(-2)$.
Solution: Calculate
$$2(-2)^2+3(-2)+1=8-6+1=3.$$
3. Define $f(x)=2x^2+3x+1$. Calculate $f(x+1)$.
Solution: Calculate
$$\begin{array}{ll}
f(x+1) &= 2(x+1)^2+3(x+1)+1 \\
&= 2(x^2+2x+1)+3(x+1)+1 \\
&= 2x^2+4x+2+3x+3+1 \\
&= 2x^2+7x+6.
\end{array}$$
4. What is the (largest) domain of $f(x)=\dfrac{1}{x-1}$ (in the real numbers)?
Solution: There are no even roots so we will focus on the denominator. We must identify the $x$'s which make the denominator zero and then remove the mfrom consideration: set the denominator equal to zero to arrive at the equation
$$x-1=0.$$
This equation has solution $x=1$. Therefore the domain is "all readl numbers except for $1$", or written in other ways: $(-\infty,1) \cup (1,\infty)$, or $-\infty \lt x \lt 1$ and $1 \lt x \lt \infty$, or $\mathbb{R} \setminus \{1\}$.
5. What is the (largest) domain of $f(x) = \dfrac{\sqrt{x-1}}{3x+2}$ (in the real numbers)?
Solution: We must consider both the root and the denominator. First we find the $x$'s that make the denominator equation to zero by setting the denominator equal to zero and solving:
$$3x+2=0,$$
so
$$3x=-2,$$
so
$$x=-\dfrac{2}{3}.$$
We will have to remove $-\dfrac{2}{3}$ from the domain. Now we must ensure that what appears under the root is non-negative (i.e. $\geq 0$):
$$x-1 \geq 0,$$
or by adding $1$,
$$x \geq 1.$$
Therefore the domain is "all real numbers greater than or equal to $1$ but not $-\dfrac{2}{3}$."
note: the statement "all real numbers greater than or equal to $1$" already excludes the point $-\dfrac{2}{3}$, so it would be ok to say just that in this circumstance
