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Mathematics 0110A/B
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Chris Brandl
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Quiz

School

Western University
Department

Mathematics

Course Code

Mathematics 0110A/B

Professor

Chris Brandl

Description

MATH 127 Fall 2012
Assignment 2 Solutions
1. For each part, sketch all of the given functions on the same set of axes.
p p p
(a) (i) f(x) = x (ii) g(x) = ▯ ▯6x (iii) h(x) = ▯2 2 ▯ x ▯ 1
5
4
3 y=f(x)
2
1
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1
y=g(x)-2 y=h(x)
-3
-4
-5
1
(b) (i) f(x) = e ▯x (ii) g(x) = ▯e ▯x+1 (iii) h(x) = + e
e ▯x
5
y=h(x)
4
3
2
y=f(x)
1
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1
-2
y=g(x)
-3
-4
-5
1 1 4
(c) (i) f(x) = (ii) g(x) = + 1 (iii) h(x) =
x x + 1 x
5
y=g(x) 4
y=h(x)
3
2
1
y=f(x)
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1
-2
-3
-4
-5
1 2. Sketch each function.
(a) f(x) = jx+1j+jx▯1j (hint: this can be expressed as a piecewise function, show
the work in getting this piecewise function)
We have that x + 1 ▯ 0 when x ▯ ▯1 and that x ▯ 1 ▯ 0 when x ▯ 1. We now
consider the three intervals x < ▯1, ▯1 ▯ x < 1, and x ▯ 1.
For the ▯rst interval x < ▯1 we have that x+1 < 0 and x▯1 < 0 so we get that
f(x) = jx + 1j + jx ▯ 1j = ▯(x + 1) ▯ (x ▯ 1) = ▯x ▯ 1 ▯ x + 1 = ▯2x.
In the second interval ▯1 ▯ x < 1 we have that x + 1 ▯ 0 and x ▯ 1 < 0 so we
get that f(x) = jx + 1j + jx ▯ 1j = (x + 1) ▯ (x ▯ 1) = x + 1 ▯ x + 1 = 2.
In the third interval x ▯ 1 we have that x + 1 ▯ 0 and x ▯ 1 ▯ 0 so we get that
f(x) = jx + 1j + jx ▯ 1j = (x + 1) + (x ▯ 1) = x + 1 + x ▯ 1 = 2x.
8
>
> ▯2x if x < ▯1
<
Thus we have that f(x) = jx + 1j + jx ▯ 1j = 2 if ▯ 1 ▯ x < 1
>
>
: 2x if x ▯ 1
4
3.5
3
2.5
2
1.5
1
0.5
-4 -3 -2 -1 0 1 2 3 4
-0.5
-1
1
(b) f(x) = (hint: ▯rst look at the graph of x ▯ 4x + 6)
x ▯ 4x + 6
2
First consider the graph of y = x ▯ 4x + 6. Completing the square we can re-
write this as y = (x ▯ 2) + 2 so we see that this is a parabola shifted 2 to the
right and 2 up. The graph of y = (x ▯ 2) + 1 goes through the point (2;2) and
approaches +1 for large negative and positive values of x. From this we see that
2 1
f(x) = goes through the point (2;1) and y approaches 0 for large
x ▯ 4x + 6 2
positive and1.5gative values of x.
1.25
1
0.75
0.5
0.25
-5 -4 -3-2 -1 0 1 2 3 4 5
-0.25
-0.5
-0.75
-1
3. For each pair of functions, ▯nd f▯g and g▯f and state the domain of each composition.
1 p
(a) f(x) = , g(x) = x
x ▯ 2
p1
We have that (f ▯g)(x) = x▯2. For x to be in the domain of f ▯g we must have
p p
that x ▯ 2 6= 0 (x 6= ▯4) AND we need x to be de▯ned (x ▯ 0). Thus, the
domain of f ▯ g is [0;4) [ (4;1).
r
1
We have that (g ▯ f)(x) = . For x to be in the domain of g ▯ f we must
x ▯ 2
have that 1 ▯ 0 AND we need 1 to be de▯ned. The ▯rst requirement gives
x▯2 x▯2
that x ▯ 2 and the second requirement gives that x 6= 2. Thus, the domain of
g ▯ f is all x > 2.
p p
(b) f(x) = x + 2, g(x) = x ▯ 4
p p
We have that (f ▯ g)(x) = x ▯ 4 + 2. For x to be in the domain of f ▯ g we
p p
must have that x ▯ 4+2 ▯ 0 AND that x ▯ 4 is de▯ned. The ▯rst requirement
is always true and the second requirement gives x ▯ 4. Thus, the domain is all
x ▯ 4.
p p
We have that (g ▯ f)(x) = x + 2 ▯ 4. For x to be in the domain of g ▯ f
p p
we must have that x + 2 ▯ 4 ▯ 0 AND we must have that x + 2 is de▯ned.
The ▯rst requirement gives that x ▯ 14 and the second requirement gives that
x ▯ ▯2. Thus, the domain is all x ▯ 14.
3 4. Give a possible equation of each graph. From observing the graph state whether the
function is even, odd, or neither.
(i) (ii)
7
3
6
5 2.5
2
4
3 1.5
1
2
1 0.5
-7 -6 -5 -4 -3 -2 --1 1 2 3 4 5 6 7 -5 -4 -3 -2 -1-0.5 1 2 3 4 5
-2 -1
-3
-1.5

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