M 115 ASSIGMENT #2 SOLUTIONS
1. Give the definition of the function.
A function is any rule that assigns to one value of the I dependent variable x one and only one value of the dependent
variable ,y.
•
•
•
•
•
•
•
•
•
X
•
•
•
•
Y
Y
X
one-to-one function
function but NOT one-to-one function
X
Y
NOT a function
2. If
f ( x ) = x + 1 and g ( x ) = 4 − x , find h( x) =
f ( x)
g ( x)
and its domain.
⇒ Domain f ( x ) : x + 1 ≥ 0, x ≥ −1,
f ( x ) = x + 1 and g ( x ) = 4 − x
Domain g ( x ) : 4 − x ≥ 0, x ≤ 4
f ( x)
g ( x)
=
this gives
x +1
4− x
, Domain: x ≥ −1 and 4 − x > 0,
x ≥ −1 ∩ x < 4
3. For the function whose equation is
⇒
f ( x) =
[ −1, 4 )
f ( x + h) − f ( x)
x
, find the average rate of change , a. r. c. =
, and
x+3
h
simplify. Leave the denominator in factored form.
( x + h )( x + 3) − x ( x + h + 3)
x+h
x
f ( x + h) − f ( x) x + h + 3 − x + 3
( x + h + 3)( x + 3)
=
⇒
h
h
h
1
2
2
x + 3 x + hx + 3h − x − hx − 3 x
3h
⇒
=
( x + h + 3)( x + 3) h
( x + h + 3)( x + 3) h
f ( x + h) − f ( x)
h
=
3
( x + h + 3)( x + 3)
M 115 A2 pg 1
4 a) Let
f ( x ) = 4 2 x 4 + 5 . Is this function odd, even or neither? Justify your answer.
f ( − x ) = 4 2(− x) 4 + 5 = 4 2 x 4 + 5 = f ( x ) ⇒ Symmetry about y axis (EVEN function)
b) Let g
( x) =
g (−x) =
c) ) Let h
x3 − x
.Is this function odd, even or neither? Justify your answer.
x4 + 1
( − x )3 − ( − x ) − x 3 + x
x3 − x
=
=
−
= − f ( x) ⇒
(− x)4 + 1
x4 + 1
x4 + 1
( x) = −
h (−x) = −
b)
c)
d)
e)
x2
.Is this function odd, even or neither? Justify your answer.
x4 + 1
x2
= h( x ) ⇒
x4 +1
EVEN function
5. For the function
a)
Symmetry about origin (ODD function)
d ( t ) = t 2 , find:
0 to 2 .
f (2) − f (0) 4 − 0
a.r.c.=
=
=2
2
2
Average rate of change from 2 to 4 .
f (4) − f (2) 16 − 4
a.r.c.=
=
= 6 = 3⋅ 2
2
2
Average rate of change from 4 to 6 .
f (6) − f (4) 36 − 16
a.r.c.=
=
= 10 = 5 ⋅ 2
6−4
2
Average rate of change from 6 to 8 .
f (8) − f (6) 64 − 36
a.r.c.=
=
= 14 = 7 ⋅ 2
8−6
2
Average rate of change from t to t + 2 .
Average rate of change from
a.r.c.=
f (t + 2) − f (t ) (t + 2) 2 − t 2 t 2 + 4t + 4 − t 2 4t + 4
=
=
=
= 2t + 2 = (t + 1) ⋅ 2
t + 2−t
2
2
2
M 115 A2 pg 2
6 Consider the piece-wise
wise function whose equation is:
2 x

f ( x ) = 1
 x2

−3≤ x < 0
x=0
x>0
a) Graph the function clearly showing the intercepts (if any).
any)
•
b) Identify the domain, range.
Domain=
[-3,+∞ ) ……………………Range= [- 6,+∞ ) ……………………….
c) Is this function one-to-one function? Why? NOT one to one as
7 Graph the function whose equation is:
f (0) = 1 = f (1)
f ( x ) = 2 − 3 x − 8 . Start with the basic function and three points on each part
and show and describe all the transformations you use to obtain the final graph. Give the equation(s) of any asymptotes.
y =−3 x →
M 115 A2 pg 3
81
8 .A page with dimensions of
2
inches by 11 inches has a border of uniform width x surrounding the printed matter
of the page (text), as shown in the diagram below.
a) Express the area A of the printed part of the page as a function of the width x of the border.
b) Give the domain and range of A.
c) If the area of the printed page is 52.46 square inches, find the width of the border.
x
x
TEXT
x
x
a)
A = (11 − 2 x )( 8.5 − 2 x ) = 4 x 2 − 39 x + 93.5
b) Domain:
0< x<
8.5
2
→
0 < x < 4.25
, Range:
0 < A < 93.5
c) 4 x 2 − 39 x + 93.5 = 52.46 ⇒ 4 x 2 − 39 x + 41.04 = 0
−b ± b 2 − 4ac 39 ±
x=
=
2a
9. If
( −39 )
2
− 4 ( 4 )( 41.04 )
2 ( 4)
=
39 ± 864.36
=8.55(reject),1.2
8
y = −2 x 2 + 5 x + 3 , graph this function and give the coordinates of the vertex, equation of axis of symmetry, find all
intercepts, domain and range, and intervals where the function is increasing and decreasing or neither.
VERTEX at xV
=−
b
5
5
5
25 25
49
=−
= + = 1.25 yV = f ( xV ) = f ( ) = −2 ⋅ + + 3 =
= 6.125
2a
2(−2)
4
4
16 4
8
INTERCEPTS: yINTERCEPT x=0 y=3;
−2 x 2 + 5 x + 3 = 0 ⇒ x =
Axis of symmetry x=x(vertex)=
x INTERCEPT y=0
−5 ± 25 + 24
⇒ x1 = 3 ; x2 = −1 / 2
−4
5
49
49
= 1.25 Maximum value=Y vertex=
= 6.125 ⇒ Range = (−∞, ]
4
8
8
Domain= R
M 115 A2 pg 4
10. A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the
rectangle. If the perimeter of the window is 20 m, what dimensions will admit the most light (maximize the area)?
semicircle
r = x
y
rectangle
2x
radius of semicircle is x, length of rectangle is y =
π ( x)
 20 − π x − 2 x 

 x , and area of semicircle is
2
2


π ( x)
 20 − π x − 2 x 
Awindow = 
 2x +
2
2


π

A =  −2 −
2

2
⇒
A=
20 − π x − 2 x
, area of rectangle is
2
2
π r2 

 , therefore the area of the window is given by:
 2 


π x2
20 x − π x 2 − x 2 π x 2
+
= −2 x 2 − π x 2 +
+ 20 x
2
2
2
b
 2
.
 x + 20 x , if this is a maximum, the max occurs at the vertex, and x = −
2a

Since this is a parabola and it opens down it is a max at the vertex, and
x=
−20
π
2.800495768, and the length is y = 10 − x − x 2.8, therefore the dimensions
π
2

2  −2 − 
2

which will maximize the area of the norman window are: length of 2.8 ft, base of 5.6 ft and radius of 2.8 ft.
11Parabolic Arch Bridge: A horizontal bridge is in the shape of a parabolic arch. Given the information shown in the
diagram, what is the height h of the arch 2 feet from the shore?
Choose the origin to be at the vertex of the parabola which gives us the equation y = ax 2 , containing the point whose
coordinates are: ( 10, -10 ), and
y = ax 2
⇒
− 10 = a (10 )
2
⇒
a=−
1
10
1 2
1
2
x and to find height at a point 2 feet in, gives an x value of 8, find y , y = − ( 8 ) = − 6.4 ,
10
10
h = 10 − 6.4 = 3.6
y=−
so
Therefore the height at a point 2 feet in from the shore is 3.6 feet.
NOTE: if you set the origin at the bottom of the parabolic arch, then the equation of the parabola becomes:
M 115 A2 pg 5
y-axis
shore
h
x=8
2 feet
x-axis
axis
2
y = ax 2 + 10, a point on it is (10, 0 ) , sub in → 0 = a (10 ) + 10
10,
1
1
, and y = − x 2 + 10, at two from the shore x = 8,
10
10
1
2
y = − ( 8 ) + 10 = 3.6 = height
10
a=−
12. Solve
x2 + 5x − 6 ≥ 0
x 2 + 5 x − 6 ≥ 0 ⇒ ( x + 6)( x − 1) ≥ 0
x
−
−
x+6
x −1
− −
− −
+
( x + 6)( x − 1) +
Answer: ( x + 6)( x − 1) ≥ 0
13. a) Graph
f ( x) =
f ( x) =
−6
+1
− 0+
+ +
+
+
− − − − − − − 0+
+
0 − − − − −
0+
for x in
+
+
+
+
+
+
+
+
+
+
( −∞, −6] ∪ [ +1, +∞ )
x+2
x−3
x + 2 x −3+3+ 2 x −3+5
5
=
=
= 1+
⇒ transformation of
x−3
x −3
x −3
x−3
f ( x) =
1
x
Stretched 5 times + translated 3 units to RIGHT + Translated 1 unit vertically
Domain= all R but x=3
f (0) = −
INTERCEPTS: Y intercept x=0
2
= −0.666...
3
X intercept y=0 x= -2
−x + 2
≠ f ( x)
−x − 3
f (− x ) ≠ − f ( x)
f (− x ) =
SYMMETRY: NONE
ASYMPTOTES Vertical asymptote:
asymptote: denominator=0 at x=3, denominator NOT zero at x=3; x=3 V.A.
Horizontal asymptote Ratio of leading coefficients =1 therefore y=1
=1 is H.A.
H.A.Oblique Asymptote NONE ( HA exists)
SIGN: use TABLE
M 115 A2 pg 6
x
+∞
−∞
-2
x+2
x−3
f ( x) =
0 +
−
−−
−
+ +
+ +
+ +
+ +
0 +
−
−−
− −
−−
+ +
+ +
+
+1 ←
0−−−−−− +
+ +
+
+ +
+ + + +
x+2
x −3
b) Graph
3
→1
||
g ( x) =
x2
x2 − 9
SYMMETRY
g (− x) =
(− x)2
x2
=
= g ( x) ⇒
(− x)2 − 9 x 2 − 9
SYMMETRY about y axis
INTERCEPTS : x-int: y=0 for x=0;
y –int: x=0 y=0
ASYMPTOTES
Vertical Asymptotes (VA):
denominator = 0 for x=3 and x=-3 and Numerator not zero for these 2 values: x=3 and x=-3 are VA
Horizontal Asymptote :
y=1 because the function approaches y=1 when x is “big”.The ratio of leading terms= 1
Oblique asymptote: None
M 115 A2 pg 7
b) Graph
h( x ) =
h( − x ) =
2x2 + 9
9
= 2 x + ⇒ y = 2 x is O. A. as
x
x
b)
d)
c)
e)
f)
g)
x + 2 y2 + 5 = 0
16. a) Ellipse, center
b)
x → ±∞
2x2 + 9
2x2 + 9
=−
= −h( x ) ⇒
−x
x
Symmetry about origin
14. a)
15. a)
9
→ 0 when
x
x2 y2
+
=1
64 9
c)
y2 x2
−
=1
9
9
3 2
 −3 5 
 −3 5  ( x + 2 )
+ ( y − 5 )2 = 1
 ,  , vertices  , ± 1
2
1
 2 2
 2 2 
4
Hyperbola, center
Asymptotes
b)
h)
(1, −2 ) , vertices (1 ± 2, −2)
( x − 1) 2 ( y + 2)2
−
=1
4
9
3
y + 2 = ± ( x − 1)
2
M 115 A2 pg 8