1001 Algebra Problems.PDF

712. c.The minimum value for both functions
     occurs at their vertex, which occurs at (0, 0).
     Also, for any positive real number a, the
     graphs of both fand gintersect the horizontal
     line y = atwice. Therefore, the range of both
     functions is [0,∞).


713. c.The radicand of an odd-indexed radical
     term (e.g., a fifth root) must be nonzero if it
     occurs in the denominator of a fraction, which
     is presently the case. As such, the restriction
     takes the form of the statement 2 – x≠0, which
     is equivalent to x≠2. Thus, the domain is
     (–∞,2)(2,∞).


714. c.The x-intercepts offare those values ofx
     satisfying the equation 1 – |2x– 1| = 0, which
     is equivalent to |2x– 1| = 1. Using the fact that
     |a| = bif and only ifa= ±b, we solve the two
     equations 2x– 1 = ±b separately:

2 x– 1 = – 1 2 x– 1 = 1

2 x= 0 2 x= 2

x= 0 x= 1

Thus, there are two x-intercepts of the given

function.

715. d.The x-values of the points of intersection of
     the graphs off(x) = x^2 and g(x) = x^4 must sat-
     isfy the equation x^4 = x^2. This equation is
     solved as follows:

x^4 = x^2

x^4 – x^2 = 0

x^2 (x^2 – 1) = 0

x^2 (x– 1)(x+ 1) = 0

x= –1, 0, 1

The points of intersection are (–1,1), (0,0),

and (1,1): there are more than two points of

intersection.

716. d.The x-values of the points of intersection of
     the graphs off(x) = 2xand g(x) = 4x^3 must sat-
     isfy the equation 4x^3 = 2x. This equation is
     solved as follows:

4 x^3 = 2x

4 x^3 – 2x= 0

2 x(x^2 – 1) = 0

4 x(x^2 – ^12 ) = 0

4 x(x– √)(x+ √) = 0

x=0, ±√)

So, the points of intersection are (0,0),

(√,2√), and (–√,–2√). There are

more than two points of intersection.

717. b.The x-values of the points of intersection of
     the graphs off(x) = ^34 x^2 and g(x) =  156 x^2 must
     satisfy the equation ^34 x^2 =  156 x^2. This equation
     is solved as follows:

^34 x^2 =  156 x^2

^34 x^2 –  156 x^2 = 0

(^34 –  156 )x^2 = 0

x^2 = 0

x= 0

Hence, there is only one point of intersection,

(0,0).

718. b.The y-intercept for a function y= f(x) is the

point (0,f(0)). Observe that f(0) =

== –2 – 24 – 0 – 44 = –1. So, the y-intercept is (0,–1).

–2 –|2 – 3(0)|

4 – 2(0)^2 |–0|

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