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(MATH021)[2008](f)midterm~cmchanaa^_10398.pdf
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HKUST
MATH021 Concise Calculus
Midterm Solution (Yellow Version)
1. Find limits
.
3x +1+ xx2 +1 . 1
(a) lim . [4 points]
x.0 x
Solution
. . .
3x +1+ xx2 +1 . 1 (3x +1+ xx2 +1 . 1)( 3x +1+ xx2 + 1 + 1)
lim = lim .
x.0 x x.0 x( x
3x +1+ x 2 + 1 + 1)
(3x +1+ xx2 + 1) . 1
= lim .
x.0
x(3x +1+ xx2 + 1 + 1)
3x + xx2 +1
= lim .
x.0 x( x
3x +1+ x 2 + 1 + 1)
3+ x2 +1 3+1
= lim . = =2 x.0 3x +1+ xx2 +1+1
1+1
....
.
(b) limln 1+sin x2 . x + 1 . [4 points]
x.1 2
Solution
........
..
limln 1+sin x2 . x +1 =ln 1+sin 12 . 1+1
x.1 22
= ln(1 + sin )
2 = ln(1 + 1)
= ln2
1
1.x
(c) lim e . [4 points]
x.1.
Solution
1
lim e 1.x = e . = x.1.
(d) lim(x . 1)2 sin 1 . [4 points]
x.1x2 . 1
Solution
From
1
.1 . sin . 1,
x2 . 1
we have
.(x . 1)2 . (x . 1)2 sin 1 . (x . 1)2 .
x2 . 1
Since
lim[.(x . 1)2] = lim(x . 1)2 =0,
x.1x.1
using the Squeeze Theorem (the Sandwich Theorem), we have
lim(x . 1)2 sin1 =0.
x.1x2 . 1
2. Find the derivatives of the following functions. sin x
(a) y = [4 points]
x + sin x
Solution
d sin xd(x + sin x)
(x + sin x) . sin x
dy dx dx
=
dx (x + sin x)2 (x + sin x) cos x . sin x(1 + cos x)
=
(x + sin x)2 x cos x . sin x
=
(x + sin x)2
.3x
(b) y =(x2 + x + 3)e[4 points]
Solution
2
dy de.3x d(x+ x + 3)
.3x
=(x2 + x +3) + e
dx dxx
2
=(x+ x + 3)(.3e.3x)+ e.3x(2x + 1) .2x
= .(3x2 + x + 8)e
1
(c) y = [4 points] 5 x2 +3x +2
Solution
dy d 1
=(x2 +3x + 2). 5
dx dx 1 1 d(x2 +3x + 2)
.1
= . (x2 +3x + 2). 5
5 dx 2x +3 6
= . (x2 +3x + 2). 5
5
(d) y =(x + 4)x2+3 [4 points]
Solution
2+3
ln y = ln(x + 4)x=(x 2 + 3) ln(x + 4)
d ln yd d(x2 + 3) d ln(x + 4)
=(x 2 + 3) ln(x +4) = ln(x + 4) + (x 2 + 3)
dxdx dx dx 1 dy x2 +3
=2x ln(x + 4) +
ydx x +4
dy x2 +3
=(x + 4)x2+3 2x ln(x + 4) +
dx x +4
2
ln(x+4)x (x2+3) ln(x+4)
Or, from y = e+3 = e,
dy (x2+3) ln(x+4) d 2+3 x2 +3
= e(x 2 + 3) ln(x +4) =(x + 4)x 2x ln(x + 4) +
dxdx x +4
3. Let u(x)= 2+ x, v(x) = ln x, and w(x)= 3+ x.
(a) Find the formula for f (x)= u(v(w(x))). [4 points]
Solution f (x)= u(v(w(x)))= 2+ v(w(x))= 2+ln w(x)= 2+ln 3+ x
(b)
Find the natural domain of f (x). [3 points] Solution The natural domain of f (x) is 3+ x> 0 or x> .3.
(c)
Find f .1(x) and f .1(3). [5 points]
Solution
y =2+ln 3+ x y . 2= ln 3+ x
y.2 2(y.2)
e= 3+ x e=3+ x
2(y.2)
x = e. 3
Thus the inverse function
2(x.2)
f .1(x)= e. 3,
and
2(3.2) 2
f .1 (3) = e. 3= e . 3.
dy d2y
4. Find and when x = 1 and y = 2, where y(x) is determined by the equation
dx dx2
23
y 3 +3xy . x = 13.
[12 points]
Solution
d 2 d(13)
(y 3 +3xy . x 3)=
dx dx 2 dy 2 dy
3y +6xy +3x . 3x 2 =0
dx dx
.
dy 3x2 . 6xy x2 . 2xy dy . 1 . 43
.
== == .
22 .
dx 3x2 +3yx2 + ydx 1+4 5
(1,2) d(x2 . 2xy) d(x2