Download Solution to Quiz 7 - MAT221 Calculus 1 - Section 1

Solution to Quiz 7 - MAT221 Calculus 1 - Section 1
1. Find the horizontal asymptotes of
f (x) =
√
4x2 + 3x − 2x
Solution: By definition, the horizontal asymptotes are found by taking the limits
of f as x → ∞ and x → −∞.
We begin by multiplying by the conjugate and then we divide by x to the highest
power of the denominator:
lim
x→∞
√
√
√
2 + 3x − 2x)( 4x2 + 3x + 2x)
(
4x
√
4x2 + 3x − 2x = lim
x→∞
4x2 + 3x + 2x
4x2 + 3x − 4x2
3x
√
= lim
= lim √
x→∞
4x2 + 3x + 2x x→∞ 4x2 + 3x + 2x
3
(3x)(x−1 )
√
√
= lim
= lim
x→∞ x−1 4x2 + 3x + 2
x→∞ ( 4x2 + 3x + 2x)(x−1 )
√
As x → ∞, we may take x > 0 and x−1 = x−2 . Therefore
lim
√
x→∞
4x2 + 3x − 2x = lim √
x→∞
3
3
3
= lim √
= .
√
4
x−2 4x2 + 3x + 2 x→∞ 4 + 3x−1 + 2
√
When finding the limit as x → −∞, we may take x < 0 and x−1 = − x−2 .
Therefore,
lim
x→−∞
√
4x2 + 3x − 2x = lim
3
3
√
= lim √
√
x→−∞ − x−2 4x2 + 3x + 2
x→∞ − 4 + 3x−1 + 2
Therefore, the limit as x → −∞ does not exist.
We have only one horizontal asymptote, y = 43 .
√
2. Sketch the graph of f (x) = x 2 + x.
Solution: The domain of f is {x | x ≥ −2}. The intercepts are (0, 0) and (−2, 0).
There are no horizonal or vertical asymptotes.
4 + 3x
f 0 (x) = √
2 2+x
f 00 (x) =
3x + 8
3
4(2 + x) 2
.
The local minimum of f is at ( −4
, f ( −4
)). f is increasing on ( −4
, ∞) and decreasing
3
3
3
on (−2, −4
).
There
are
no
inflection
points.
f
is
never
concave
down
and is concave
3
up on (−2, ∞).