Calculus I Midterm Nov. 18, 2011
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Guidelines for the test:
• Put your name or student ID number on every page.
• There are 11 problems.
• The exam is closed book; calculators are not allowed.
• There is no partial credit for problems 1.
• For problems 2-10, please show all work, unless instructed otherwise. Partial credit will be given only for work shown. Print as legibly as possible - correct answers may have points taken off, if they’re illegible.
• Mark the final answer.
1. (5 pts) Assume that f0(x0) = f00(x0) = 0, f000(x0) > 0. Answer the following True/False questions (True ⇒ ; False ⇒ ×).
• × f0(x0) is a local maximum value of f0(x).
• f0(x0) is a local minimum value of f0(x).
• × f (x0) is a local maximum value of f (x).
• × f (x0) is a local minimum value of f (x).
• the point (x0, f (x0)) is an inflection point of the curve y = f (x).
Sol. Since f1 00(x0) = 0 which says that x0 is the critical number of f0(x), to- gether with f000(x0) > 0, by 2ndderivative test, x0 is the local minimum of f0(x).
f2 000(x0) > 0 says that, the rate of change of f00(x) at x = x0 is positive.
By the fact that f00(x0) = 0, we have f00(x) < 0, x ∈ (x0 − δ, x0) (or f (x) is concave downward at the left side of x0) and f00(x0) > 0, x ∈ (x0, x0+ δ) (or f (x) is concave upward at the right side of x0) , for some δ > 0. Thus we can conclude that x0 is the inflection point of f (x).
2. Evaluate the limits.
(a) (5 pts) lim
x→∞
√x sin(x1).
Sol. Define y = 1x, x > 0. Then lim
x→+∞
√x sin(1x) = lim
y→0+
q1
y sin y = lim
y→0+
sin y y ·√
y
= 1 · 0 = 0.
(b) (5 pts) lim
x→2
(x2−x−2)10 (x2−10x+16)10. Sol. lim
x→2
(x2−x−2)10
(x2−10x+16)10 = lim
x→2
[(x−2)(x+1)]10
[(x−2)(x−8)]10 = lim
x→2
(x−2)10(x+1)10
(x−2)10(x−8)10 = lim
x→2 (x+1)10 (x−8)10 =
310
(−6)10 = 2110 = 10241 . 3. (5 pts) Given that lim
x→a
f (x)−f (a)
(x−a)2 = −1, does f0(a) exist? If f0(a) exist, compute f0(a). If not, prove it.
Sol. f0(a) = lim
x→a
f (x)−f (a)
x−a = lim
x→a
f (x)−f (a)
(x−a)2 · (x − a).
Since lim
x→a
f (x)−f (a)
(x−a)2 = −1 (which says that lim
x→a
f (x)−f (a)
(x−a)2 exists), and lim
x→a(x − a) = 0 (which says that lim
x→a(x − a) exists), we have that f0(a) = lim
x→a
f (x) − f (a)
(x − a)2 ·(x−a) = lim
x→a
f (x) − f (a) (x − a)2 ·lim
x→a(x−a) = (−1)·0 = 0.
Thus, f0(a) exists and f0(a) = 0.
4. (5 pts) Use a linear approximation to estimate √ 104.
Sol. Define f (x) =√
x, then we have f (x0+ h) ≈ f (x0) + f0(x0)h be the linear approximation of f (x) at x = x0+ h. Let x0 = 100 and h = 4, by the fact that f0(x) = 2√1x, we have
f (104) ≈ f (100) + f0(100) · 4 = 10 + 1
20 · 4 = 10.2.
5. (a) (5 pts) d
dx(x2− 1 x2)10. Sol. d
dx(x2− 1
x2)10= 10(x2− 1 x2)9· d
dx(x2− 1
x2) = 10(x2− 1
x2)9· (2x + 2 x3).
(b) (5 pts) d
dx[sin2x + sin (x3)].
Sol. d
dxsin2x + sin(x3) = 2 sin x · d
dx(sin x) + cos(x3) · d dx(x3)
= 2 sin x · cos x + cos(x3) · 3x2 = sin 2x + 3x2cos(x3).
6. Given x2+ y2 = 1,
(a) (5 pts) find dydx implicitly;
Sol.
x2+ y2 = 1
⇒ d
dx(x2+ y2) = d dx(1)
⇒ 2x + 2y · d dxy = 0
⇒ dy
dx = −x y. (b) (5 pts) find ddx2y2 implicitly.
Sol. By (a), we have dydx = −xy. Then
d2y
dx2 = dxd dydx
= dxd
−xy
= −y·
d
dx(x)−x·dxd(y) y2
= −y−x·(−xy)
y2
= −y2y+x3 2
= −y13. (since x2 + y2 = 1)
7. (5 pts) Given f (x) = x(x + 1)(x + 2)(x + 3) . . . (x + 100), find f0(0).
Sol.
f0(0) = lim
h→0
f (h)−f (0) h−0
= lim
h→0
h(h+1)(h+2)···(h+100)−0 h−0
= lim
h→0(h + 1)(h + 2) · · · (h + 100)
= 1 · 2 · 3 · · · 100
= 100!.
8. (a) (5 pts) Given f (x) = 2x2+ 1, find a function F (x) such that F0(x) = f (x).
Sol. F (x) = 23x3+ x + c, where c is a real number.
(b) (5 pts) Given g(x) = sin(2x), find a function G(x) such that G0(x) = g(x).
Sol. G(x) = −12cos(2x) + c, where c is a real number.
9. (10 pts) Find the points on the ellipse 4x2+ y2 = 4 that are farthest away from the point (6, 0).
Sol. To find the points on the ellipse that are farthest away from (6,0), that is, to maximize p(x − 6)2+ (y − 0)2 subject to 4x2+ y2 = 4, we sufficiently to solve
max (x − 6)2 + y2
s.t. 4x2+ y2 = 4. (1)
Since 4x2 + y2 = 4, we can rewrite the constraint of (1) as y2 = 4 − 4x2, where −1 ≤ x ≤ 1. That is, the problem (1) can be converted as
−1≤x≤1max (x − 6)2+ (4 − 4x)2 = max
−1≤x≤1−3(x + 2)2 + 52 .
To find the critical number of the function f (x) = −3(x + 2)2+ 52, since f0(x) = −6(x + 2), and x0 = −2 does not located on the domain [-1,1], there is no critical number.
Since f (−1) = 49 and f (1) = 25, by Close Interval Method, the maximum value of f (x) is 49 as x = −1.
Thus, the farthest point away from (6,0) on the ellipse is (-1,0) with the distance√
49 = 7.
10. (10 pts) Car A is traveling west at 90 km/h and car B is traveling north at 100 km/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 60 m and car B is 80 m from the intersection (兩車以多快的相對速度接近)?
Sol. The distance between car A and car B is z(t) where z(t)2 = x(t)2+ y(t)2. Since z(t)2 = x(t)2+ y(t)2, we have
2z(t)d
dtz(t) = 2x(t)d
dtx(t) + 2y(t)d
dty(t). (2)
By the hypothesis, we have d
dtx(t) = 90 km/h; d
dty(t) = 100 km/h;
x(t) = 0.06 km; and y(t) = 0.08 km, then we can compute z(t) = px(t)2+ y(t)2 = 0.1 km, by (2), we have
d
dtz(t) = x(t)dtdx(t) + y(t)dtdy(t)
z(t) = 0.06 · 90 + 0.08 · 100
0.1 = 134 km/h.
11. (total 20 points; (a)-(n) no partial credit) Study the function f (x) = x2 x2+ 4 and answer the following questions.
(a) (1 pt) Domain of f : Df = R.
(Because f (x) can be defined everywhere on R.) (b) (1 pt) Horizontal Asymptote: y = 1 .
(Since lim
x→+∞
x2
x2+4 = lim
x→+∞
1 1+ 4
x2
= 1 and lim
x→−∞
x2
x2+4 = lim
x→−∞
1 1+ 4
x2
= 1.) (c) (1 pt) Vertical Asymptote: none .
(Because the denominator of f (x) always positive.) (d) (1 pt) f0(x) = (x28x+4)2 .
(since f (x) = 1 − x24+4, we have f0(x) = (x28x+4)2.)
(e) (1 pt) Interval(s) of increasing of f : [0, +∞) or (0, +∞) . (f0(x) is positive in this interval.)
(f) (1 pt) Interval(s) of decreasing of f : (−∞, 0) or (−∞, 0] . (f0(x) is negative in this interval.)
(g) (1 pt) Local maxima of f : none .
(h) (1 pt) Local minima of f : local minimum value of f (x) is 0 at the point x0 = 0 .
(by 1st derivative test.) (i) (1 pt) f00(x) = −8(3x(x2+4)2−4)3 .
(f00(x) = dxdf0(x) = (x2+4)2·8−8x·2·(x(x2+4)4 2+4)·2x = 8(x2(x+4)−32x2+4)3 2 = −8(3x(x2+4)2−4)3 .) (j) (1 pt) Interval(s) of concave up: (−√23,√23).
(f00(x) is positive in this interval.)
(k) (1 pt) Interval(s) of concave down: (−∞, −√2
3)S(√23, +∞) . (f00(x) is negative in these interval.)
(l) (1 pt) Inflection point(s) of f : (√2
3,14) and (−√2
3,14) . (Since the left side of x = −√2
3 is concave downward, and the right side of x = −√2
3 is concave upward, thus x = −√2
3 is a inflection point. Similar argument to x = √23.)
(m) (1 pt) x-intercepts of f : 0 .
(x−intercepts takes value as y = 0) (n) (1 pt) y-intercepts of f : 0 .
(y−intercepts takes value as x = 0)
(o) (6 pts) Sketch the graph of f showing all significant features.
