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Section 1.4 Additional Exercises

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 \(y=-\frac{1}{3}x+\frac{13}{3}\)


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 \(\dfrac{x^2-1}{3x-1}\leq 1\text{.}\)



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 \(x\text{.}\)


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.



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



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



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 \(\cos\theta=\cos\frac{38\pi}{5}\text{.}\)



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 \((1,-1)\text{.}\)



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 \(y\text{?}\)



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

\begin{equation*} \frac{x^2+3x-5}{x+2}=Q(x)+\frac{R}{x+2}\text{.} \end{equation*}

\(Q(x)=x+1\text{,}\) \(R=-7\)