1001 Algebra Problems.PDF

719. b.If there is no x-value that satisfies the equa-
     tion,f(x) = 0 then the graph ofy= f(x) does
     not cross the horizontal line y= 0, which is the
     x-axis.


720. b.If there is no x-value that satisfies the equa-
     tion,f(x) = 3 then there is no point on the
     graph ofy= f(x) when y= 3. Therefore, 3 is
     not in the range off.

Set 46 (Page 109)

721. c.The domain of a rational function is the set
     of all real numbers that do not make the
     denominator equal to zero. For this function,
     the values ofxthat must be excluded from the
     domain are the solutions of the equation x^3 – 4x
     = 0. Factoring the left side yields the equivalent
     equation

x^3 – 4x= x(x^2 – 4) = x(x– 2)(x+ 2) = 0

The solutions of this equation are x= –2, 0,

and 2. Hence, the domain is (–∞,–2)∪( –2,

0)∪(0, 2)∪(2,∞).

722. d.First, simplify the expression for f(x) as
     follows:

= =

=

While there is a hole in the graph offat x = 4,

there is no x-value that makes the denomina-

tor of the simplified expression equal to zero.

Hence, there is no vertical asymptote. But, since

the degrees of the numerator and denomina-

tor are equal, there is a horizontal asymptote

given by y= 1 (since the quotient of the coeffi-

cients of the terms of highest degree in the

numerator and denominator is 11 = 1).

723. b.We must identify the intervals in the domain
     ofp(x) on which the graph ofy= p(x) rises
     from left to right. This happens on the intervals
     (–3, 0)∪( 2,∞).


724. b.The graph offpasses the horizontal line test
     on this interval, so, it has an inverse.


725. c.Determining the inverse function for f
     requires that we solve for xin the expression
     y= :

y=  5 xx–+^12

y(5x+ 2) = x– 1

5 xy+ 2y= x– 1

5 xy– x= –2y– 1

x(5y– 1) = –2y– 1

x= – 52 yy–– 11

Now, we conclude that the function f–1(y) =

– 52 yy–– 11 ,y≠^15 is the inverse function off.

726. d.Remember that the domain offis equal to
     the range off–1. As such, since 0 does not
     belong to the range off–1, it does not belong to
     the domain off, so f(0) is undefined. Also, the
     fact that (1, 4) is on the graph ofy= f(x) means
     that f(1) = 4; this is equivalent to saying that
     (4, 1) is on the graph off–1, or that f–1(4) = 1.
     All three statements are true.


727. b.Determining the inverse function for f
     requires that we solve for xin the expression
     y= x^3 + 2:

y= x^3 + 2

y– 2 = x^3

x= ^3 y– 2

Hence, the inverse is f–1(y) = ^3 y– 2.

x– 1

5 x+ 2

x^2 + x– 12)

x^2 + 9

(x– 3)(x+ 4)

(x^2 + 9)

(x– 3)(x– 4)(x+ 4)

(x^2 + 9)(x– 4)

(x– 3)(x^2 – 16)

(x^2 + 9)(x– 4)

ANSWERS & EXPLANATIONS–