MA 3972-MA-Book May 8, 2018 13:46Review of Precalculus 83- There is one relative minimum point
at (3,−27). Thus, the relative minimum
value for the function is−27. There is no
relative maximum. (See Figure 5.8-7.)
[−2, 5] by [−30, 10]
Figure 5.8-75.9 Solutions to Cumulative Review Problems
- The notationf(2)=4 means that when
x=2,y=4, and thus, the point (2, 4) is
on the graph of f(x).Similarly,
f(−4)=10 implies that the point
(−4, 10) is also on the graph. Since f(x)
is a linear function, its graph is a line. The
slope of a line,m, is defined as
m=
y 2 −y 1
x 2 −x 1
. Thus,m=
10 − 4
− 4 − 2
=
6
− 6
=− 1.
Using the point slope of a line
y−y 1 =m(x−x 1 ), you have
y− 4 =−1(x−2) ory=−x+6.- Enter into your calculatory 1 =x^3 −xand
examine the graph. (See Figure 5.9-1.)
[−3, 3] by [−3, 3]
Figure 5.9-1
Note thatf(x)≥0 on the intervals
[−1, 0] and [1,∞). Therefore, the
solution tox^3 −x≥0is− 1 ≤x≤0or
x≥1.- Sincef(x)=
1
x,
f(x+h)− f(x)
h=
1
x+h−1
x
hand the
lowest common denominator (LCD) of
1
x+h
and1
x
isx(x+h). Multiplying the
numerator and denominator of the
complex fraction by the LCD, you have1
x+h−1
x
h·
x(x+h)
x(x+h)
, which is equivalentto
x−(x+h)
xh(x+h)
or
−h
xh(x+h)
or− 1
x(x+h).
- Begin by findingg−^1 (x). Rewrite
g(x)= 3 x−12 asy= 3 x−12. Switchx
andy, and you havex= 3 y−12. Solving
fory, you havey=
x+ 12
3
. Substitute
g−^1 (x) fory. Thus,g−^1 (x)=
x+ 12
3
andg−^1 (3)=3 + 12
3
=5.
- The slope of the line segment joining the
origin (0, 0) and the point of tangency
(4,−3) ism=
− 3 − 0
4 − 0
=
− 3
4
. Since this line
segment is perpendicular to the tangent
line, the slope of the tangent line is
4
3
.
Using the point-slope form of a line, you
havey−y 1 =m(x−x 1 ), or
y−(−3)=4
3
(x−4) ory=4
3
x−25
3
. (See
Figure 5.9-2.)
[−14.5, 14.5] by [−7, +7]
Figure 5.9-2