Problem 1 Consider the functions f(x) = x

2 � 1 and g(x) =

px� 1.

(a) Find the domains of functions f and g. Give your answers in interval nota-

tion:

Domain of f : This function does not have a denominator to consider, and

contain a root (square or otherwise), so the domain includes all real num-

bers. In interval notation: (�1,1).

Domain of g: This function contains a square root, so we need to be sure

that the portion under the square root is greater than or equal to zero:

x� 1 � 0! x � 1.

In interval notation: [1,1).

(b) Find h(x) = (f � g)(x). Then give the domain of h in interval notation.

h(x) = f(g(x)) = f(

px� 1) =

�px� 1

�2 � 1 = x� 1� 1 = x� 2.

The domain of a composite function is the set of x for which the inner

function (in this case g(x)) is defined and the set of x for which h(x) are

defined. In this case, the domain for g(x) is [1,1), and the x for which

h(x) is defined (�1,1), so the domain of h(x) is:

[1,1).

(c) Which function, f or g, has a greater average rate of change from x = 2 to

x = 5? Justify your answer.

The average rate of change for any function f is calculated as ARC =

f(x2)�f(x1)x2�x1

.

ARC for f :

ARCf(x) =

f(5)� f(2)

5� 2

=

5

2 � 1� (2

2 � 1)

3

=

24� 3

3

= 7.

ARC for g:

ARCg(x) =

g(5)� g(2)

5� 2

=

p5� 1�

p2� 1

3

=

2� 1

3

= 1/3.

In the calculations above, the ARC for f is greater than the ARC for g, so

f has the larger average rate of change between x = 2 and x = 5.

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(d) Consider a function m(x) = 2f(x). Prove that from x = a to x = b, the

average rate of change of function m is two times the average rate of change

of function f.

ARC for f :

f(b)� f(a)

b� a

.

ARC for m:

m(b)�m(a)

b� a

=

2f(b)� 2f(a)

b� a

=

2 (f(b)� f(a))

b� a

= 2

✓f(b)� f(a)

b� a

◆.

The above shows that for any two points a and b, the average rate of change

for m(x) is twice that of f(x).

2

Given the equation f(x) =

1

2

x

�1/2(3x+ 4)

1/2 � 3

2

x

1/2(3x+ 4)

�1/2

1. Find the greatest common factor of f(x).

2. Factor the equation completely.

3. Evaluate f(2).

4. Solve f(x) = 0. Is there are no solutions, state “none”.

Solutions:

1. The greatest common factor is

1

2

x

�1/2(3x+ 4)

�1/2.

2. Factoring the equation we have

f(x) =

1

2

x

�1/2(3x+ 4)

�1/2((3x+ 4)� 3x)

=

1

2

x

�1/2(3x+ 4)

�1/2(4)

= 2x

�1/2(3x+ 4)

�1/2

=

2

px

p3x+ 4

3. f(2) =

2p2

p3(2) + 4

=

2p2

p10

=

1p5

4. There are no solutions because 2 6= 0.

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