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TEST 1 SOLUTIONS
1.5pts Find the natural domain and range of f ( x ) = x 2 − 3
Since x 2 ≥ 3; x ≥ 3 or x ≤ − 3 is the domain. Since
2.5pts Explain why the function
a)
b)
x 2 − 3 ≥ 0; y ≥ 0 is the range.
( x + 2 ) ( x2 − 1)
has one or more holes in its graph and
f ( x) =
( x + 2 )( x − 1)
state the x-values at which those holes occur,
The function is not defined if x = -2 or x = 1
find a function g whose graph is identical to that of f , but without the holes.
( x + 2 ) ( x 2 − 1) ( x + 2 )( x + 1)( x − 1)
f ( x) =
=
= x + 1 if we ignore the holes –
( x + 2 )( x − 1)
( x + 2 )( x − 1)
so g ( x ) = x + 1
3.8pts Sketch the graph of
1
by
y = 2−
x +1
translating and reflecting
1
the graph of y = .
x
Graph intercepts and show
any asymptotes as broken
lines.
Note that the intercepts
Are (0, 1) and (-1/2, 0)
4.9pts Let f ( x ) = x and g ( x ) = x3 + 1 . Find the following:
a)
b)
c)
f ( g ( 2)) = f (9) = 9 = 3
g ( f ( 4)) = g ( 2) = 9
f ( f (16 ) ) = f ( 4 ) = 2
1
5.5pts Sketch the graph of
y = 2 ( x + 1)
by making appropriate
transformations to the
graph of a basic power
function. Be sure to
graph intercept[s].
2
1
as a composition of two functions g and h so that f = g h
x−3
1
The “inner” function is x – 3, “outer” is 1/x, so g ( x ) = , h ( x ) = x − 3
x
7.7pts Express f ( x ) = x + 3x + 1 in piecewise form without using absolute values. Simplify.
6.5pts Express f ( x ) =
 x + 3 x + 1 , x ≥ 1 4 x + 1 , x ≥ 1
f ( x ) = x + 3x + 1 = 
=
 − x + 3 x + 1, x < 1 2 x + 1 , x < 1
8.10pts Find formulas for f g and g f and state the domains of the compositions:
f ( x ) = x − 3 ; g ( x ) = x 2 + 3 Simplify. Note that for f, the domain is x ≤ 3 and
for g it is all real #’s
f  g ( x )  =
x 2 + 3 − 3 which is like problem 1, but here x 2 + 3 − 3 ≥ 0 , x 2 + 3 ≥ 3
and since both are positive, we square both sides to get x 2 + 3 ≥ 9 ; x 2 ≥ 6 ; x ≥ 6 or x ≤ − 6 for
this domain
g  f ( x )  = g
(
) (
x −3 =
x −3
)
2
+ 3 = x − 3 + 3 = x which looks like the domain should be
x ≥ 0 but look at the fact that it came from x − 3 so the domain is x ≥ 3
9.10pts True-False:
a)
The domain of a logarithmic function is the interval x > 1. FALSE [x > 0]
b)
The range of the absolute value function is all positive real numbers. FALSE [all
non-negative numbers – includes 0
c)
The domain of f + g is the intersection of the domains of f and g . TRUE
d)
If f and g are inverse functions then they have the same domains.FALSE
e)
If f ( a ) = L , then lim f ( x ) = L . FALSE
x→a
2

 3 
10.5pts Find the exact value of sec sin −1  −  
 4 

3
 3
Let sin −1  −  = x , sin x = − which is in QIV by range of sin −1 . This
4
 4
would have been critical if, e.g. you’d been asked for tan instead of sec. However, here
since sin x = b/r and a > 0 in QIV, all you have to do is find a:
a 2 + b 2 = r 2 ; a 2 = 16 − 9 = 7 so a = 7 is one way to get this [or you could use appropriate
4
identities]. Since sec x = r/a, it is
7
11.6pts
Find the exact value of
1
 1 
a)
log 2   = x ; 2 x =
= 2−5 ; x = −5
32
 32 
b)
ln
( e ) = ln e
1/ 2
= 1/ 2
Find a formula for f −1 ( x ) : f ( x ) = 5 4 x + 2
12.5pts
1 5
x − 2 ) = f −1 ( x )
(
4
13.10pts
Solve for x without using a calculator:
1
a)
log10 x = −1 ; 10−1 = x ; x = 10−2 =
or 0.01
100
3
1

1
b)
ln   + ln ( 2 x 3 ) = ln 3 ; ln  ⋅ 2 x 32  = ln 3 ; ln ( 2 x 2 ) = ln 3; 2 x 2 = 3; x = ±
2
x
x

BUT the negative makes the first term given undefined, so we only get
3
x=
2
EXTRA CREDIT
14.
Find a formula for f −1 ( x ) and find the domain and range of f −1 : f ( x ) = 4 x 2 + 3, x ≤ 0
y = 5 4 x + 2 ; x = 5 4 y + 2 ; x 5 = 4 y + 2; y =
f ( x ) = 4 x 2 + 3, x ≤ 0 ; y = 4 x 2 + 3, x ≤ 0 becomes x = 4 y 2 + 3, y ≤ 0 so we have
x −3 1
=
x − 3 = f −1 ( x ) [We have to choose the negative because of the
4
2
constraint on y that came from the given constrain on x.]
Hence the range of f −1 is y ≤ 0 and looking at the formula for it. the domain is x ≥ 3 [and if
you look carefully at the given function, you’ll see that the range of it is y ≥ 3
y=−
15.
16.
State the transformations necessary to change the graph of y = x into y = 2 − x − 3
Reflect given graph in the y-axis, stretch vertically times 2 and shift down 3
1
Find the domain and range of the function f ( x ) =
1 + cos x
The domain is all values of x such that cos x ≠ −1 ; so x ≠ π + 2kπ k any integer.
The range is like the one we did in class:
1
1
−1 ≤ cos x ≤ 1 ; 0 ≤ 1 + cos x ≤ 2; 1 + cos x ≤ 2 gives us
≥ which is the range.
1 + cos x 2
3
4