- b.While x= 1 satisfies the original equation,x
= –1 cannot, because negative inputs into a
logarithm are not allowed.
- a.The denominator in the quadratic formula
is 2a, which in this case is 2, not 1. The com-
plex solutions should be x=i 5 .
- b.The signs used to define the binomials on
the right side should be switched. The correct
factorization is (x– 7)(x+ 3).
- a.You must move all terms to one side of the
inequality, factor (if possible), determine the
values that make the factored expression equal
to zero, and construct a sign chart to solve
such an inequality. The correct solution set
should be [–2, 2].
- a.The left side must be expanded by FOILing.
The correct statement should be (x– y)^2 =
x^2 – 2xy+ y^2.
- b.This equation has no real solutions because
the output of an even-indexed radical must be
nonnegative.
- c.There is no error.
- b.The left side is not a difference of squares. It
cannot be factored further.
- b.You cannot cancel terms of a sum in the
numerator and denominator. You can cancel
only factors common to both. The complex
fraction must first be simplified before any can-
cellation can occur. The correct statement is:
= = = =
=
- b.The first equality is incorrect because the
natural logarithm of a sum is not the sum of
the natural logarithms. In fact, the expression
on the extreme left side of the string of equali-
ties cannot be simplified further.
975. a.The first equality is incorrect because 2 log 5
(5x) = log 5 (5x)^2 = log 5 (25x^2 ). The other equal-
ities are correct as written.
976. c.There is no error.
Set 63 (Page 148)
- b.The line y= 0 is the horizontal asymptote
for f.
- a.The line is vertical, so its slope is undefined.
- a.The point is actually in Quadrant II.
- b.The domain offmust be restricted to [0,∞)
in order for fto have an inverse. In such case,
the given function f–1(x) = xis indeed its
inverse.
- c.There is no error.
- c.There is no error.
- c.There is no error.
- a.The coordinates of the point that is known
to lie on the graph ofy= f(x) are reversed;
they should be (2,5).
- c.There is no error.
- b.-2 is not in the domain ofg, so that the
composition is not defined at –2.
- a.The point (0,1) is the y-intercept, not the
x-intercept, off.
- a.The graph ofg is actually decreasing as x
moves from left to right through the domain.
- a.The graph ofy= f(x+ 3) is actually obtained
by shifting the graph ofy= f(x) to the left 3
units.
- c.There is no error.
- a.You cannot distribute a function across
parts of a single input. As such, the correct
statement should bef(x– h) = (x– h)^4.
- a.The graph ofy= 5 passes the vertical line
test, so it represents a function. It is, however,
not invertible.
^2 y– x
y+ 4x
xy^
y + 4 x
^2 y– x
xy
^2 yx–yx
y+xy^4 x
^2 xyy– xxy
xyy+ ^4 xxy
^2 x– ^1 y
1
x+ ^4 y
^2 x–1– y–1
x–1+ 4y–1
ANSWERS & EXPLANATIONS–
