Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 1

Brief Review of Algebra and Trigonometry for Calculus

Finishing This Unit

1. Review the objectives of this unit and make sure you are able to meet all of them.
2. If there is a concept, definition, example or exercise that is not yet clear to you, go back and reread it, then contact your tutor for help.
3. Definitions are important, for each definition you should be able to give an example (something that satisfies the definition) and a counterexample (something does not satisfy the definition).
4. Do the exercises in “Learning from Mistakes” section for this unit.

Learning from Mistakes

There are mistakes in each of the following solutions. Identify the errors and give the correct answer.

1. If ${\displaystyle x}$ is any number in the interval ${\displaystyle \left[4,\infty \right)}$, in which interval is ${\displaystyle 1-\frac{x}{2}}$?

   Erroneous Solution

   If ${\displaystyle x}$ is in ${\displaystyle \left[4,\infty \right)}$, then ${\displaystyle 4\le x}$, and ${\displaystyle -4\le -x}$, thus ${\displaystyle -2\le \frac{-x}{2}}$.

   Therefore, ${\displaystyle -1\le 1-\frac{x}{2}}$, and we conclude that ${\displaystyle 1-\frac{x}{2}}$ is in the interval ${\displaystyle \left[-1,\infty \right)\text{.}}$

2. Simplify the following expressions.

   1. ${\displaystyle {\left(a+3b\right)}^{2}a{b}^{-1}}$

   2. ${\displaystyle {\left(x^3{y}^{-3}\right)}^2}$

   3. ${\displaystyle \left(a{b}^2-b{c}^3\right)abc}$

   Erroneous Solutions

   1. ${\displaystyle {\left(a+3b\right)}^{2}a{b}^{-1}=\left(a^2+9b^2\right)a{b}^{-1}=a^3{b}^{-1}+9ab=\frac{a^3}{b}+9ab}$

   2. ${\displaystyle {\left(x^3{y}^{-3}\right)}^2=x^5{y}^{-1}=\frac{x^5}{y}}$

   3. ${\displaystyle \left(a{b}^2-b{c}^3\right)abc=a{b}^2-a{b}^2{c}^3}$

3. Factor the following expressions.

   1. ${\displaystyle 2x^2-7x-4}$

   2. ${\displaystyle 3x^5+24x^2}$

   3. ${\displaystyle 81x^{4}y-y^5}$

   Erroneous Solutions

   1. By the quadratic formula,

      ${\displaystyle x=\frac{7\pm \sqrt{49-4\left(2\right)\left(-4\right)}}{4}=\frac{7\pm 9}{4};}$

      hence, the solutions are

      ${\displaystyle x=4}$ and ${\displaystyle x=\frac{-1}{2}}$,

      and the factors are

      ${\displaystyle 2x^2-7x-4=\left(x-4\right)\left(x+\frac{1}{2}\right).}$

   2. ${\displaystyle \begin{array}{ll}3x^5+24x^2\hfill & =3x^2(x^3+8)\hfill \\ \hfill & =3x^2(x+2)(x^2+xy+4)\hfill \end{array}}$

   3. ${\displaystyle \begin{array}{ll}81x^{4}y-y^5\hfill & =(9x^2\sqrt{y}-y^{5/2})(9x^2\sqrt{y}+y^{5/2})\hfill \\ \hfill & =\sqrt{y}(9x^2+y^2)\hfill \\ \hfill & =\sqrt{y}(3x-y)(3x+y)(9x^2+y^2)\hfill \end{array}}$

4. 1. Simplify the expression ${\displaystyle \frac{x^2+9x+20}{x^2+5x}\text{.}}$

   2. Rationalize the expression ${\displaystyle \frac{\sqrt{4+h}-2}{h}\text{.}}$

   Erroneous Solutions

   1. ${\displaystyle \frac{x^2+9x+20}{x^2+5x}=\frac{9x+20}{5x}}$

   2. ${\displaystyle \frac{\sqrt{4+h}-2}{h}=\frac{2+\sqrt{h}-2}{h}=\frac{\sqrt{h}}{h}+\frac{1}{\sqrt{h}}}$

5. Give the exact values of the following trigonometric functions

   1. ${\displaystyle \text{sec }\left(\frac{7\pi }{6}\right)}$

   2. ${\displaystyle \text{sin }\left(\frac{7\pi }{12}\right)}$

   Erroneous Solutions

   1. ${\displaystyle \text{sec}\left(\frac{7\pi }{6}\right)=\frac{1}{\text{cos}\left({\displaystyle \frac{7\pi }{6}}\right)}=\frac{1}{\text{cos}\left({\displaystyle \frac{\pi }{6}}\right)}=2}$

   2. ${\displaystyle \begin{array}{rl}\sin \left({\displaystyle \frac{7\mathrm{\pi <!--\; \pi \; -->}}{12}}\right)& =\sin ({\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{3}}+{\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{4}})\\ & =\sin \left({\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{3}}\right)+\sin \left({\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{4}}\right)\\ & ={\displaystyle \frac{\sqrt{3}}{2}}+{\displaystyle \frac{\sqrt{2}}{2}}\\ & ={\displaystyle \frac{\sqrt{5}}{2}}\end{array}}$