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

Question 1

i

$ax=a\,$$ax=a\,$

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

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

$1\cdot x=1$$1\cdot x=1$

$x=1\,$$x=1\,$

ii

$(x-y)(x+y)=(x\cdot (x-y)+y\cdot (x-y))$$(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}-xy+xy-y^{2}\,$$=x^{2}-xy+xy-y^{2}\,$

$=x^{2}-y^{2}\,$$=x^{2}-y^{2}\,$

iii

$x^{2}=y^{2}\,$$x^{2}=y^{2}\,$, then

$x^{2}-y^{2}=(x+y)(x-y)=0\,$$x^{2}-y^{2}=(x+y)(x-y)=0\,$

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

iv

$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})\,$$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})\,$

$=(x-y)x^{2}+(x-y)xy+(x-y)y^{2}\,$$=(x-y)x^{2}+(x-y)xy+(x-y)y^{2}\,$

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

$=x^{3}-y^{3}\,$$=x^{3}-y^{3}\,$

v

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

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

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

$=x^{n}-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\,$$x^{2}=xy\,$ implies that $x=y\,$$x=y\,$ and thus $x-y=0\,$$x-y=0\,$. Step 4 requires division by $x-y=0\,$$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}\,$$-3<x^{2}\,$, and $x^{2}>=0\,$$x^{2}>=0\,$ for all x.

iii

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

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

v

$x^{2}-2x+2$$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\,$$(x^{2}-2x+1)+1>0\,$

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

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

vi

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

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

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

viii

Complete the square:

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

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

ix

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

x

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

xi

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

$log_{2}2^{x}<log_{2}8=x<3\,$$log_{2}2^{x}<log_{2}8=x<3\,$

xii

$x<1\,$$x<1\,$

xiii

NOTE: The answer in the 3rd Edition provides $0<x<1$$0<x<1$ or $x>1$$x>1$. Plugging in values $x>1$$x>1$, e.g. 10, gives us $1/10-1/9<0$$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$${\frac {1}{x}}+{\frac {1}{1-x}}>0$, then ${\frac {1}{x-x^{2}}}>0$${\frac {1}{x-x^{2}}}>0$, which is only positive when $0<x<1$$0<x<1$ since $x^{2}>x$$x^{2}>x$.

Question 5

i

$a+c<b+d\,$$a+c<b+d\,$

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

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

ii

$b-a>0\,$$b-a>0\,$

$=-a+b>0\,$$=-a+b>0\,$

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

$=-b<-a\,$$=-b<-a\,$

iv

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

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

v

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

$a(1-a)>0\,$$a(1-a)>0\,$

$=a^{2}-a>0\,$$=a^{2}-a>0\,$

$=a^{2}>a\,$$=a^{2}>a\,$

viii

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

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

ix

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

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