Solutions

  1. If a = 5 then defining b = a2 produces b = 25. Now define a = $\sqrt{{%
2}}$. The value of b is now b = 2.

  2. Evaluate followed by Simplify yields
    $\displaystyle {\frac{{f(x+h)-f(x)}}{{h}}}$ = $\displaystyle {\frac{{\left( x+h\right) ^{2}+3h-x^{2}}}{{h}}}$  
      = 2x + h + 3  

    Select the expression = ${\frac{{\left( x+h\right) ^{2}+3h-x^{2}}}{{h}}}$ and with the CTRL key down drag the expression to create a copy. Select the expression $\left(\vphantom{ x+h}\right.$x + h$\left.\vphantom{ x+h}\right)^{{2}}_{}$ and with the CTRL key down choose Expand. Add similar steps (use Factor to rewrite 2xh + h2 + 3h) until you have the following:
    $\displaystyle {\frac{{f(x+h)-f(x)}}{{h}}}$ = $\displaystyle {\frac{{\left( x+h\right) ^{2}+3h-x^{2}}}{{h}}}$  
      = $\displaystyle {\frac{{x^{2}+2xh+h^{2}+3h-x^{2}}}{{h}}}$  
      = $\displaystyle {\frac{{2xh+h^{2}+3h}}{{h}}}$  
      = $\displaystyle {\frac{{h\left( 2x+h+3\right) }}{{h}}}$  
      = 2x + h + 3  

  3. To rewrite f (x) = max$\left(\vphantom{ x^{2}-1,7-x^{2}}\right.$x2 -1, 7 - x2$\left.\vphantom{ x^{2}-1,7-x^{2}}\right)$ as a piecewise-defined function, first note that the equation x2 -1 = 7 - x2 has the solutions x = - 2 and x = 2. The function f is given by

    g(x) = $\displaystyle \left\{\vphantom{
\begin{array}{lll}
x^{2}-1 & \text{if} & x<-...
...ext{if} & -2\leq x\leq 2 \\
x^{2}-1 & \text{if} & x>2
\end{array}
}\right.$$\displaystyle \begin{array}{lll}
x^{2}-1 & \text{if} & x<-2 \\
7-x^{2} & \text{if} & -2\leq x\leq 2 \\
x^{2}-1 & \text{if} & x>2
\end{array}$

    As a check, note that f (- 5) = 24, g(- 5) = 24, f (1) = 6, g(1) = 6, f (3) = 8, and g(3) = 8.

  4. Construct the following table:

    n $\varphi$(n) n $\varphi$(n) n $\varphi$(n) 1 1 11 10 21 12 2 1 12 4 22 10 3 2 13 12 23 22 4 2 14 6 24 8 5 4 15 8 25 20 6 2 16 8 26 12 7 6 17 16 27 18 8 4 18 6 28 12 9 6 19 18 29 28 10 4 20 8 30 8              

    Notice, for example, that
    $\displaystyle \varphi$(4⋅5) = 8 = $\displaystyle \varphi$(4)$\displaystyle \varphi$(5)  
    $\displaystyle \varphi$(4⋅7) = 12 = $\displaystyle \varphi$(4)$\displaystyle \varphi$(7)  
    $\displaystyle \varphi$(3⋅8) = 8 = $\displaystyle \varphi$(3)$\displaystyle \varphi$(8)  

  5. We have the following table:

    n d (n) n d (n) n d (n) 1          

    Notice that d (n) consists of all the divisors of n.



Subsections