These problems require a comprehensive knowledge of the skills reviewed in this chapter. They are not in any particular order. A proficiency in these skills will help you a long way as your learn the calculus material in the following chapters.

Rationalize the denominator for each of the following expressions. That is, re-write the expression in such a way that no square roots appear in the denominator. Also, simplify your answers if possible.

1. $\dfrac{1}{\sqrt{2}}$

2. $\dfrac{3h}{\sqrt{x+h+1}-\sqrt{x+1}}$

1. $\dfrac{\sqrt{2}}{2}$

2. $3(\sqrt{x+h+1}+\sqrt{x+1})$

Solve the following equations.

1. $2-5(x-3)=4-10x$

2. $2x^2-5x=3$

3. $x^2-x-3=0$

4. $x^2+x+3=0$

5. $\sqrt{x^2+9}=2x$

1. $-13/5$

2. $-1/2,3$

3. $(1\pm\sqrt{13})/2$

4. No real solutions

5. $\sqrt{3}$

By means of counter-examples, show why it is wrong to say that the following equations hold for all real numbers for which the expressions are defined.

1. $(x-2)^2=x^2-2^2$

2. $\dfrac{1}{x+h}=\dfrac{1}{x}+\dfrac{1}{h}$

3. $\sqrt{x^2+y^2}=x+y$

Counter-examples may vary.

1. $x=3$

2. $x=h=1$

3. $x=y=1$

Find an equation of the line passing through the point $(-2,5)$ and parallel to the line $x+3y-2=0\text{.}$

$x+3y-13=0\text{,}$ or equivalents such as $y=-\frac{1}{3}x+\frac{13}{3}$

Solution

We first put the equation of the given line into standard form:

\begin{equation*} \begin{gathered} x + 3y -2 = 0 \\ 3y = -x + 2 \\ y = -\frac{1}{3}x + \frac{2}{3} \end{gathered} \end{equation*}

Thus, the slope of this line is $-\frac{1}{3}\text{.}$ The line we seek needs to be parallel to this line, and so it must be of the form

\begin{equation*} y = -\frac{1}{3}x + b\text{.} \end{equation*}

To find $b\text{,}$ we use the fact that the line passes through the point $(-2,5)\text{.}$ That is, it satisfies

\begin{equation*} 5 = - \frac{1}{3} (-2) + b\text{.} \end{equation*}

From this, we see that $b = \frac{13}{3}\text{.}$ Hence, the equation of the line is

\begin{equation*} y = -\frac{1}{3} x + \frac{13}{3}\text{.} \end{equation*}

Solve $\dfrac{x^2-1}{3x-1}\leq 1\text{.}$

$(-\infty,0]\cup(\frac{1}{3},3]$

Explain why the following expression never represents a real number (for any real number $x$): $\sqrt{x-2}+\sqrt{1-x}\text{.}$

It is impossible for both $x-2$ and $1-x$ to be non-negative for the same real number $x\text{.}$

Solution

The domain of $\sqrt{x-2}$ is $x \geq 2\text{.}$ The domain of $\sqrt{1-x}$ is $x \leq 1\text{.}$ Since there exist no $x \in \mathbb{R}$ which lie in both domains, the expression $\sqrt{x-2}+\sqrt{1-x}$ does not ever represent a real number.

In other words, the domain of $\sqrt{x-2}+\sqrt{1-x}$ is the intersection of $\{x \in \mathbb{R} | x \geq 2\}$ and $\{x \in \mathbb{R} | x \leq 1\}\text{,}$ which is the empty set, $\emptyset\text{.}$

Simplify the expression $\dfrac{\left[3(x+h)^2+4\right]-\left[3x^2+4\right]}{h}$ as much as possible.

$6x+3h$

Simplify the expression $\dfrac{\frac{x+h}{2(x+h)-1}-\frac{x}{2x-1}}{h}$ as much as possible.

$-1/\left[(2x+2h-1)(2x-1)\right]$

Simplify the expression $-\sin x(\cos x+3\sin x)-\cos x(-\sin x+3\cos x)\text{.}$

$-3$

Solve the equation $\cos x=\frac{\sqrt{3}}{2}$ on the interval $0\leq x\leq 2\pi\text{.}$

$\pi /6\text{,}$ $5\pi /6$

Find an angle $\theta$ such that $0\leq\theta\leq\pi$ and $\cos\theta=\cos\frac{38\pi}{5}\text{.}$

$2\pi/5$

What can you say about $\dfrac{\left\vert x\right\vert+\left\vert 4-x\right\vert}{x-2}$ when $x$ is a large (positive) number?

It is equal to 2 for all $x$ larger than 4.

Find an equation of the circle with centre in $(-2,3)$ and passing through the point $(1,-1)\text{.}$

$(x+2)^2+(y-3)^2=25\text{.}$

Find the centre and radius of the circle described by $x^2+y^2+6x-4y+12=3\text{.}$

Centre is $(-3,2)$ and radius is 2.

If $y=9x^2+6x+7\text{,}$ find all possible values of $y\text{.}$

$y$ could be any real number greater than or equal to 6.

Simplify $\left(\dfrac{3x^2 y^3 z^{-1}}{18x^{-1}yz^3}\right)^2\text{.}$

$x^6 y^4/(36z^8)$

If $y=\dfrac{3x+2}{1-4x}\text{,}$ then what is $x$ in terms of $y\text{?}$

$x=(y-2)/(3+4y)$
Divide $x^2+3x-5$ by $x+2$ to obtain the quotient and the remainder. Equivalently, find polynomial $Q(x)$ and constant $R$ such that
$Q(x)=x+1\text{,}$ $R=-7$