MAC 1147 (3032) - Quiz #3
Name:
Solution Key
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1
Find f ◦ g and g ◦ f . (2 points)
f (x) =
√
3
x − 5,
(f ◦ g)(x) = f (g(x))
= f (x3 + 1)
p
= 3 (x3 + 1) − 5
p
3
= x3 − 4.
2
g(x) = x3 + 1.
(g ◦ f )(x) = g(f (x))
√
= g( 3 x − 5)
√
3
= 3x−5 +1
=x−5+1
= x − 4.
Find the inverse of f . State the domain and range of both f and f −1 . (3 points)
8x − 4
f (x) =
2x + 6
8y − 4
x=
2y + 6
x(2y + 6) = 8y − 4
2xy + 6x = 8y − 4
2xy − 8y = −6x − 4
y(2x − 8) = −6x − 4
−6x − 4
−2(3x + 2)
3x + 2
y=
=
=−
.
2x − 8
2(x − 4)
x−4
Dom(f ) = Rng(f −1 ) = {x 6= −3}.
Rng(f −1 ) = Dom(f ) = {x 6= 4}.
3
The path of a diver is given by
4
24
y = − x2 + x + 12
9
9
where y is the height (in feet) and x is the horizontal distance from the end of the diving board (in
feet). What is the maximum height of the diver? (3 points)
b
24/9
24/9
=−
=
= 3.
2a
2(−4/9)
8/9
4
24
Maximum height = − (3)2 + (3) + 12 = −4 + 8 + 12 = 16.
9
9
x-value of the vertex = −
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