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

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Question 1[edit]

i[edit]

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[edit]

(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[edit]

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[edit]

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[edit]

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[edit]

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


Question 2[edit]

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[edit]

ii[edit]

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


iii[edit]

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


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


v[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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


xi[edit]

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


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


xii[edit]

x < 1\,


xiii[edit]

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[edit]

i[edit]

a+c<b+d\,


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

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


ii[edit]

b-a>0\,


=-a+b >0\,

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

=-b < -a\,


iv[edit]

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

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


v[edit]

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

a(1-a) > 0\,

= a^2 - a > 0\,

=a^2 > a\,


viii[edit]

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[edit]

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[edit]

ii[edit]

If 0 \leq x < y then x^n < y^n\, by 6.i.


If x < 0 \leq y then x^n < 0\, and y^n \geq 0, hence also true.


If x < y < 0\, then 0 < -y < -x\,, which means (-y)^n < (-x)^n\,. Therefore, -(-y)^n > -(-x)^n = y^n > x^n\,.


iii[edit]

We already know that if x < y then x^n < y^n. If x^n < y^n or y^n < x^n then x < y or y < x. But since x^n = y^n, x = y.


iv[edit]

We already proved in 6.i that if 0 < x < y, then x^n < y^n for all n \in \mathbb{N}^+.

Thus, if x^n = y^n it cannot be x < y or y < x, and must be x = y. If x = -y, then x^n = (-y)^n, which is positive by virtue of n being even.