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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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