MA 3972-MA-Book April 11, 2018 12:11

26 STEP 2. Determine Your Test Readiness

3.4 Solutions to Diagnostic Test

Chapter 5

1. Write 5x− 2 y=10 in slope-intercept form
   y=mx+band obtainy=

5

2

x−5. The slope

of the liney=

5

2

x−5is

5

2

. Therefore, the

slope of the line perpendicular toy=

5

2

x− 5

is−

2

5

. Since the line perpendicular to

y=

5

2

x−5 also passes through the origin, its

y-intercept is 0 and its equation isy =−

2

5

x.

2. Using your calculator, sety 1 = 3 | 2 x− 4 |and
   y 2 =6. Examine the figure below and note
   that 3| 2 x− 4 |>6 whenx<1orx>3.
   Thus, the solution in interval notation is
   (−∞,1)∪(3,∞).

[−3, 7] by [−3, 10]

3. 
   

   Sincef(x)=x^2 +1,f(3)=10 and
   f(3+h)=(3+h)^2 + 1 = 9 + 6 h+h^2 + 1 =
   10 + 6 h+h^2.
   The quotient
   f(3+h)−f(3)
   h

   
   

   10 + 6 h+h^2 − 10
   h

   
   

=

6 h+h^2

h

= 6 +h.

4. Simplify the equation 4 lnx− 2 =6 and
   obtain lnx=2. (Note thatexand lnxare
   inverse functions.) Thus,elnx=e^2 orx=e^2.


5. Sinceg(x)= 2 x−2,g(2)=2(2)− 2 =2.
   Therefore, f(g(2))=f(2). Alsof(x)=x^3 −8.
   Thus,f(2)= 23 − 8 =0.

Chapter 6

6. See the figure below.
   Ifb=2, thenx=−1 would be a solution for
   f(x)=2.
   Ifb=0, or−2, f(x)=2 would have two
   solutions.
   Thus,b=3, choice (A).

− 2

− 1

− 2 − 1 0

1

4

2

3

(−2, 4) (0, 4)

y

x

y = 2

7. xlim→−∞

√

x^2 − 4

2 x

=xlim→−∞

√

x^2 − 4

/

(−

√

x^2 )

2 x

/

(−

√

x^2 )

(Note: asx→−∞,x=−

√

x^2 .)

=x→lim−∞

−

√

(x^2 −4)

/

x^2

2

=x→lim−∞

−

√

1 −(4/x^2 )

2

=−

√

1

2

=−

1

2

8. h(x)=

{√

x ifx> 4

x^2 −12 ifx≤ 4

xlim→ 4 +h(x)=xlim→ 4 +

√

x=

√

4 = 2

lim

x→ 4 −

h(x)=lim

x→ 4 −

(x^2 −12)=(4^2 −12)= 4

Since limx→ 4 +h(x) =xlim→ 4 −h(x), thus limx→ 4 h(x)

does not exist.