Download Exam 1 solutions

Calculus I
Math 141 Fall 2009
Professor Ben Richert
Exam 1
Solutions
Problem 1. (10pts) Where is the function f (x) =
(
x2 + x + 6
sin x
x<0
continuous?
x≥0
Solution. Recall that f (x) is continuous at x = a if lim f (x) = f (a) which itself occurs if and only if
x→a
lim+ f (x) = f (a) = lim− f (x). Of course, we know that x2 + x + 6 and sin x are continuous on all R, so
x→a
x→a
it follows that f (x) is continuous on all x except possibly x = 0 (the only place where the right and left
hand limits might differ). Observe that lim f (x) = lim sin x = 0, since f (x) = sin x for x > 0 and, as
x→0+
x→0+
already mentioned, sin x is continuous, so we may compute the limit by evaluating at zero. On the other
hand, observe that lim f (x) = lim (x2 + x + 6) = 6, since f (x) = x2 + x + 6 for x < 0 and again, as
x→0−
x→0−
already mentioned, x2 + x + 6 is continuous, so we evaluate at zero to compute the limit. Since the right
and left hand limits aren’t equal at zero, lim f (x) does not exist, and thus does not equal f (0) = sin 0 = 0.
x→0
So we conclude that f (x) is not continuous at 0, and thus is continuous on (−∞, 0) ∪ (0, ∞).
Problem 2. (10pts) Use the limit definition of the derivative to compute f ′ (x) if f (x) = 1/x.
Solution. By the definition of the derivative, we have that
f (x + h) − f (x)
= lim
h→0
h→0
h
f ′ (x) = lim
1
x+h
−
1
x
x
x(x+h)
−
h
.
Simplifying yields
lim
h→0
1
x+h
−
1
x
h
= lim
1 x
x+h x
h
h→0
x−x−h
x(x+h)
−
1 x+h
x x+h
= lim
h→0
x+h
x(x+h)
h
−h
−1
= lim 2
.
h→0
h
x(x + h)h
x + xh
−1
A limit law now allows us to evaluate the quotient at h = 0 (because 2
is a rational function in h
x + xh
whose denominator is not zero at h = 0). We obtain
= lim
h→0
= lim
h→0
lim
h→0 x2
so f ′ (x) =
−1
−1
= 2,
+ xh
x
−1
.
x2
sec(2x)
. Boxes are not absolutely required, but you must
(x + 1) sin x
show all your work. Just writing down the end result will not earn full credit. Be sure and say, for instance,
which rules you are using and where you are using them. A few phrases should suffice.
Problem 3. (10pts) Compute f ′ (x) if f (x) =
Solution. I’ll use boxes. We use the the quotient rule to differentiate f (x) =
gf ′ − f g ′
d f
.
=
dx g
g2
sec(2x)
:
(x + 1) sin x
f = sec(2x)
g = (x + 1) sin x
f ′ = 2 sec(2x) tan(2x) g ′ = sin x + (x + 1) cos x
yielding
((x + 1) sin x)(2 sec(2x) tan(2x)) − sec(2x)(sin x + (x + 1) cos x)
.
((x + 1) sin x)2
I filled in the bottom two boxes of the table above using . . . two more boxes. First, the chain rule to
differentiate sec(2x)
d
f (g(x) = f ′ (g(x))g ′ (x)
dx
f = sec(x)
f ′ = sec x tan x
so
g = 2x
g′ = 2
d
sec(2x) = 2 sec(2x) tan(2x), and then the product rule to differentiate (x + 1) sin x
dx
d
f g = f ′g + f g′
dx
f = (x + 1) g = sin x
f′ = 1
g ′ = cos x
yielding
d
(x + 1) sin x = sin x + (x + 1) cos x.
dx
Problem 4. (15pts) The trail of a snail on the loose on your dorm room floor can be described by the
equation
y 2 sin(x) = x sin(y 2 ).
(The origin is at the center of the room, the y-axis points due north, and x and y are in feet).
√
dy
at the point (π, π). (The number may be a bit ... gross; don’t simplify).
(a–10pts) Compute
dx
Solution. We use implicit differentiation, remembering that near most points, y is a function of
x. So
d
d 2
y sin x =
x sin(y 2 )
dx
dx
2yy ′ sin x + y 2 cos(x) = sin(y 2 ) + x cos(y 2 )2yy ′
2yy ′ sin x − 2xy cos(y 2 )y ′
′
2
y (2y sin x − 2xy cos(y ))
y′
= sin(y 2 ) − y 2 cos(x)
= sin(y 2 ) − y 2 cos(x)
sin(y 2 ) − y 2 cos(x)
=
2y sin x − 2xy cos(y 2 )
√
At the point (π, π), we obtain:
√
√
dy 0 − π(−1)
π
1
sin(( π)2 ) − ( π)2 cos(π)
√
√
√
√ = √ .
=
=
= √
dx (π,√π)
2 π sin π − 2π π cos(( π)2 )
0 − 2π π(−1)
2π π
2 π
In the calculation above, we wrote that
d
y = y′
dx
using the chain rule,
d 2
y = 2yy ′
dx
using the chain rule,
using the product rule,
using the chain rule, and
d 2
y sin x = 2yy ′ sin x + y 2 cos x
dx
d
sin(y 2 ) = cos(y 2 )2yy ′
dx
d
x sin(y 2 ) = sin(y 2 ) + x cos(y 2 )2yy ′
dx
using the product rule.
(b–5pts) What is the meaning of the number you found in part (a)? Answer with a sentence or two.
Solution. This is the rate of change of the y coordinate of the snail with respect to the x coordinate.
1
So, instantaneously, how many feet he moves to the north ( √ feet) per foot he moves to the
2 π
east.
Problem 5. (10pts) What values of a and b ensure that the function f (x) = ax2 + 2x + b has tangent
y = 4x + 1 at x = 1?
Solution. Recall that the derivative of f (x) at x = 1 gives the slope of the tangent, and additionally, that
the tangent line to f (x) at x = 1 touches the graph of f (x) at x = 1. We can read the slope of the tangent
off the equation y = 4x + 1 (its 4). Also, we can evaluate the line at x = 1 to find the point on the curve
(y = 4(1) + 1 = 5, so (1, 5) is on the curve). So, we have f ′ (1) = 4 and f (1) = 5. Of course, f ′ (x) = 2ax + 2,
so 4 = f ′ (1) = 2a + 2 and a = 1. Then f (x) = x2 + 2x + b and 5 = f (1) = 1 + 2 + b implies that b = 2. Problem 6. (10pts) Consider the following graph, then match each of the entries in the left column with one
entry from the right column. Use each entry from the right column exactly once. This is the only problem
on the exam for which you need show no work and for which no English is required.
C
A
lim f (x)
x→1+
lim
x→1
f (x) − f (1)
x−1
(A) −1
(B) 1/2
(C) 1
B
δ > 0 such that if |x − 2| < δ, then |f (x) − 2| < 1
(D) 3/2
H
ǫ > 0 such that if |x − 2| < 1, then |f (x) − 2| < ǫ
(E) 2
E
L such that |f (x) − L| < 1 whenever 0 < |x − 5| < 1/4
(F) 7/2
G
a such that |f (x) − 4| < 2 whenever 0 < |x − a| < 1/2
D
F
(G) 4
L such that f (x) can be made arbitrarily close to L by taking x sufficiently close to 5/2.
(H) 6
a such that f (x) can be made arbitrarily close to 5 by taking x sufficiently close to a.