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

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

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

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

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

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

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

$$= x^n - y^n\,$$

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 $$x^2+x-1 >0\,$$.

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

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

viii
Complete the square:

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

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

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:

$$log_2 2^x < log_2 8 = x < 3\,$$

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
$$a+c0\,$$

$$= (b-a)+(d-c) > 0\,$$, 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 $$a^2 < b^2$$.

=Question 6=