Brief Solutions for Calculus Quiz #3 1. f (x) = |1 − |x||, −1 ≤ x ≤ 2.
- 1 1 2
1
X
f(x) has a local maximum 1 at x = 0, 2.
f(x) has a local minimum 0 at x = 1, −1.
And the global maximum and minimum are the same with local maximum and minimum respectively.
2. (a) f (x) =√
x+ 1, x ≥ −1.
f′(x) = 12(x + 1)−12 >0, it is increasing.
f′′(x) = −14(x + 1)−32 <0, it is concave down.
f(x) has a local minimum 0 at x = −1.
(b) f (x) = (1 + x)−2, x6= −1
f′(x) = −2(1 + x)−3 = > 0 , if x ∈ (−∞, −1) incerasing
<0 , if x ∈ (−1, ∞) decreasing f′′(x) = 6(1 + x)−4 >0, it is concave up.
limx→±∞f(x) = 0, limx→−1f(x) = ∞. Thus y = 0, x = −1 are asymptotes.
(c) f (x) = xe−x, x≥ 0.
f′(x) = e−x(1 − x) =
>0 , if x ∈ (0, 1) incerasing 0 , x= 1
<0 , if x ∈ (1, ∞) decreasing
f′′(x) = e−x(x − 2) =
<0 , if x ∈ (0, 2) concave down 0 , x= 2 inflection point
>0 , if x ∈ (2, ∞) concave up
f′′(1) < 0, f (x) has a local maximum f (1) = e−1. And it has a local minimum 0 at x= 0.
limx→∞ exx = 0, thus y = 0 is a asymptote.
(d) f (x) = ln x + 1x, x >0
f′(x) = x−1− x−2 =
<0 , if x ∈ (0, 1) decerasing 0 , x= 1
>0 , if x ∈ (1, ∞) increasing
f′′(x) = −x−2+ 2x−3 =
>0 , if x ∈ (0, 2) concave up 0 , x= 2 inflection point
<0 , if x ∈ (2, ∞) concave down f′′(1) = 1 > 0, f (x) has a local minimum f (1) = 1.
limx→0+f(x) = ∞, thus x = 0 is a asymptote.
3.
B(3) − B(0) = B′(t)(3 − 0)
⇒ |B′(t)| = |B(3) − 3 3 | ≤ 1
⇒ 0 ≤ B(3) ≤ 6 4. (a) A = 2 = 12r2θ ⇒ θ = r42
We want to minimize rθ + 2r.
Let f (r) = rr42 + 2r = 4r + 2r.
Solve f′(r) = −4r−2+ 2 = 0 ⇒ r =√
2, θ = 2 f′′(r) = 8r−3 >0 for all r > 0. Thus f (√
2) is a local minimum.
(b) Similarly to part (a), we can find that r =√
10, θ = 2 5. (a) Let y = limx→∞(1 + x3)x.
ln y = lim
x→∞
ln(1 + 3x)
1 x
= lim
t→0
ln(1 + 3t) t
= lim
t→0 3 1+3t
1
= 3 Thus limx→∞(1 + 3x)x = e3.
(b)
lim
x→0+( 1 sin x − 1
x) = lim
x→0+
x− sin x xsin x
= lim
x→0+
1 − cos x sin x + x cos x
= lim
x→0+
sin x 2 cos x − x sin x
= 0
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(c)
lim
x→π
2−
1tan x = lim
x→π
2−
exp[tan x ln 1]
= exp[ lim
x→π
2
−
tan x ln 1]
= e0
= 1 (d)
lim
x→π
2−
(sin x)tan x = lim
x→π
2−
exp[tan x ln(sin x)]
= exp[ lim
x→π
2
−
tan x ln(sin x)]
= exp[ lim
x→π
2
−
ln(sin x)
1 tan x
]
= exp[ lim
x→π
2−
cos x sin x
− tan−2xsec2x]
= exp[ lim
x→π
2−− cos x sin x]
= e0
= 1 6. (a) Let f (x) = 0.5+xx .
(b) Solve x = 0.5+xx .
⇒ x = 0, 0.5.
f′(x) = 0.5+x−x(0.5+x)2 = (0.5+x)0.5 2.
Then f′(0) = 2 > 1, x∗ = 0 is unstable.
|f′(0.5)| = 0.5 < 1, x∗ = 0.5 is locally stable.
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