MATH (Tonight, Easy, Pre-calc)
jusinc1
sample_notation.pdf
Math 111 Worksheet 1 Solutions Show ALL work. Solutions must satisfy the requirements specified in the Math 111 DOCUMENTATION GUIDELINES.
1) Using interval notation, state the domain for 5
( ) 2
x f x
x +
= −
. Provide rationale for your solution.
The domain of f is all reals such that the radicand in the numerator is greater than or equal to zero,
, and the denominator cannot equal zero, 5 0x + ≥ 2 0x − ≠ .
5 0 5x x+ ≥ ⇒ ≥ − and 2 0 2x x− ≠ ⇒ ≠
-5 2-5 2 The domain of f is [ 5 , 2) (2, )− ∞∪
2) Algebraically evaluate ( )( )f g x− as a single fraction in unfactored form given 3( ) 2 1 x
f x x −
= +
and
2 ( )
4 x
g x x −
= +
( )( )
( )
( )
2 2
2
2 2
2
2 2
2
( ) ( ) 3 2
2 1 4 3 4 2 2
2 1 4 4 2 1 ( 3)( 4) (2 1)( 2)
(2 1)( 4)
3 4 12 2 4 2 2 8 4
12 2 3 2 2 9 4 12 2 3 2
2 9
f g x f x g x x x x x x x x x x x x x
x x x x x x
x x x x x x x x x
x x x x x x
x x x x x x
− = −
− − = −
+ + − + − +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= ⋅ − ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ − + − + −
= + +
− + − − − + − =
+ + + + − − − −
= + +
+ − − + + =
+ +
1
2
2
4 4 10
2 9 4 x x x x
− + − =
+ +
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3) Algebraically evaluate h
xfhxf )()( −+ given xxxf 32)( 2 −=
( ) ( ) ( )
( )
( )
2 2
2 2
2 2 2
2
2 3 2( ) ( )
2 2 3 3
2 4 2 3 2
4 2 3
4 2 3
4 2 3, 0
x h x h xf x h f x h h
2
3
2 3
x
x xh h x h x h
x xh h h x h
xh h h h
h x h h
x h h
+ − + − −+ − =
+ + − − − =
+ + − − =
+ − =
+ − =
= + − ≠
x+
4) Let 2 4
( ) 2
x x f x
x +
= −
. Algebraically solve ( ) 1f x =
( )( )
2
2
2
( ) 1 4
1 2 4 2
3 2 0 2 1 0
1, 2
f x x x x
x x x x x
x x x
=
+ =
− + = −
+ + =
+ + =
= − −
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5) If a projectile is shot into the air, it attains a maximum height and falls back to the ground. Suppose that seconds corresponds to the time when the projectile’s height is at its maximum of 0t = 400h = feet. Then if air resistance is ignored, the height of the projectile can be modeled by: where 2( ) 16 400h f t t= = − + h represents the height of the projectile above the ground (feet) t represents time since the projectile was at its maximum height (seconds) Evaluate and explain its meaning in practical terms. ( 2)f −
( )2( 16 2 400 336
f − − +2)− = =
At 2 seconds before the projectile was at its maximum height, the projectile was 336 feet above the ground.
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