Download MATH 114 W09 Quiz 1 Solutions 1 1. a) Find the domain of the

MATH 114 W09 Quiz 1 Solutions
1
x
.
3x + 1
it Solution: We must have 3x + 1 > 0. So x > − 13 . The domain is all values x
where x > − 31 .
√
b) Find the range of the function f (x) = x2 − 4 + 1.
2
Solution: The domain is |x| ≥ 2. For
√ these value of x, the function x − 4 can
be any nonnegative √
value. As such, x2 − 4 can be any nonnegative value. This
means that f (x) = x2 − 4 + 1 can be any value y where y ≥ 1. So the range of
f is y ≥ 1.
3|x| + x
c) Find the domain and range of the function
.
x
Solution: The only restriction is that the denominator in the fraction cannot be
zero. So the range is all numbers x except x = 0. When x > 0, where have
= 4x
= 4. When x < 0, we have |x| = −x, and
|x| = x, and thus f (x) = 3x+x
x
x
−3x+x
−2x
thus f (x) = x = x = −2. So the range of f are the values 4 and −2.

 x + 2, if x < −1;
−x,
if −1 ≤ x < 1;
2. Sketch the graph of the function f (x) =

x − 1, if 1 ≤ x.
1.
a) Find the domain of the function f (x) = √
2
MATH 114 W09 Quiz 1 Solutions
3. The graph of f is given below. Use it to find the graph of g(x) = − 21 f (x − 1).
Solution: To get the graph of g(x), we take the graph of f and first shift it right one
unit. The we perform a vertical compression by a factor 21 . Lastly, we reflect the curve
about the x-axis.
2
1.5
1
0.5
0
1
2
3
x
4
5
6
–0.5
–1
4. For the functions f (x) = x1 , and g(x) =
g ◦ f.
)=
Solution: f ◦ g(x) = f (g(x)) = f ( x+1
x+2
g ◦ f (x) = g(f (x)) = g( x1 ) =
1
+1
x
1
+2
x
x+1
x+2
1
x+1
x+2
find f ◦ g and g ◦ f. Find the domain of
=
x+2
.
x+1
. The function is defined when both x 6= 0 (since f is
only defined for x 6= 0) and x1 + 2 6= 0. If x1 + 2 = 0, then
the domain of g ◦ f is all numbers x except x = 0, − 12 .
1
x
= −2, or x = − 12 . Therefore
Bonus Suppose f and g are odd functions. What can one say about f ◦ g?
Solution: f ◦g is odd. To see this, we have f ◦g(−x) = f (g(−x)) = f (−g(x)) (since g is odd).
Also, f (−g(x)) = −f (g(x)) = −f ◦ g(x) (since f is odd). Therefore, f ◦ g(−x) = −f ◦ g(x)
and f ◦ g is odd.