Solutions To Mathematics Textbooks/Calculus (3rd) (0521867444)/Chapter 1

Question 1

i

$ax=a$

$a\cdot (x\cdot a^{-1})=a\cdot a^{-1}=1$

$a\cdot (a^{-1}\cdot x)=1$

$(a\cdot a^{-1})\cdot x)=1$

$1\cdot x=1$

$x=1$

ii

$(x-y)(x+y)=(x\cdot (x-y)+y\cdot (x-y))$

$=(x^2-xy)+(xy-y^2)$

$=x^2-xy+xy-y^2$

$=x^2-y^2$

iii

$x^2=y^2$, then

$x^2-y^2=(x+y)(x-y)=0$

Either $(x+y)=0$, which means $x=-y$ or $(x-y)=0$, which means that $x=y$.

iv

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$=(x-y)x^2+(x-y)xy+(x-y)y^2$

$=(x^3-x^{2}y+x^{2}y-x{y}^2+x{y}^2-y^3$

$=x^3-y^3$

v

$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+x{y}^{n-2}+y^{n-1})$

$=x(x^{n-1}+x^{n-2}y+...+x{y}^{n-2}+y^{n-1})-y(x^{n-1}+x^{n-2}y+...+x{y}^{n-2}+y^{n-1})$

$=(x^n+x^{n-1}y+...+x^2{y}^{n-2}+x{y}^{n-1})-(x^{n-1}y+x^{n-2}{y}^2+...+x{y}^{n-1}+y^n)$

$=x^n-y^n$

Note that the two middle terms do no appear to match, but if you follow the logical development in each term, you will see that there is a matching additive inverse for each term on both sides.

vi

Use the same method as iv, expand the expression and cancel.

Question 2

$x^2=xy$ implies that $x=y$ and thus $x-y=0$. Step 4 requires division by $x-y=0$ and thus is an invalid step.

Question 4

ii

All values of x satisfy the inequality, since it can be rewritten as $-3<x^2$, and $x^2>=0$ for all x.

iii

If $5-x^2<-2$, then $x^2>7$.

Thus, $x>\sqrt{7}$, or $x<-\sqrt{7}$.

v

$x^2-2x+2$ cannot be factored in its current form, so we first turn a part of the expression into a perfect square:

$(x^2-2x+1)+1>0$

We then complete the square on $x^2-2x+1=(x-1{)}^2$, so now we have the expression:

$(x-1{)}^2+1>0$, which is positive for all values of x.

vi

If $x^2+x+1>2$, then $x^2+x-1>0$.

$x^2+x-1=(x+\frac{1+\sqrt{5}}{2})(x-\frac{\sqrt{5}-1}{2})$

Thus $x<-\frac{1+\sqrt{5}}{2}$, or $x>\frac{\sqrt{5}-1}{2}$.

viii

Complete the square:

$x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}={(x+\frac{1}{2})}^2+\frac{3}{4}$

Thus, ${(x+\frac{1}{2})}^2+\frac{3}{4}>0$.

ix

Solve for $(x-\pi )(x+5)>0$ first, then consider the third factor $(x-3)$ for both cases, giving us $-5<x<3$, or $x>\pi$

x

$x>\sqrt{2}$, or $x<\sqrt[3]{2}$

xi

If $2^x<8$, then taking the base 2 logarithm on both sides:

$lo{g}_2{2}^x<lo{g}_{2}8=x<3$

xii

$x<1$

xiii

NOTE: The answer in the 3rd Edition provides $0<x<1$ or $x>1$. Plugging in values $x>1$, e.g. 10, gives us $1/10-1/9<0$, so I think this is a misprint or an incorrect answer.

If $\frac{1}{x}+\frac{1}{1-x}>0$, then $\frac{1}{x-x^2}>0$, which is only positive when $0<x<1$ since $x^2>x$.

Question 5

i

$a+c<b+d$

$=(b+d)-(a+c)>0$

$=(b-a)+(d-c)>0$, which is true since $(b-a),(d-c)>0$

ii

$b-a>0$

$=-a+b>0$

$=-a-(-b)>0$

$=-b<-a$

iv

$(b-a),c>0$, therefore $c(b-a)>0$.

$c(b-a)=bc-ac$, therefore $ac<bc$.

vi

$a>1$, therefore $1-a>0$.

$a(1-a)>0$

$=a^2-a>0$

$=a^2>a$

viii

If $a$ or $c$ are 0, then by definition $ac<bd$.

Otherwise, we have already proved that $ac<bc$ for $a,c>0$, therefore if $c<d$, $ac<bd$ is true.

ix

If $a=0$, then $a^2<b^2$ since $b^2>0$.

If $0<a<b$, then $a^2<ab$. Since $ab<b^2$, we have $a^2<b^2$.