Download Exam 3, Solutions

NAME:
Calculus with Analytic Geometry I
Exam 3, Friday, September 16, 2011
3 pages, 4 exercises, one bonus exercise, 110 points
1. (15 points) For what value of the constant c is the function f continuous on (−∞, ∞)? Explain why the
function is continuous for your choice of c. This explanation should not take up more than one, at
most two, sentences.
{
cx2 + 2x if x < 2
f (x) =
x3 − cx if x ≥ 2
Solution.
(Exercise 45 of §2.5-Similar to Exercise 46 that was done in class) We have
lim f (x) = lim− (cx2 + 2x) = 4c + 4;
x→2−
x→2
lim f (x) = lim+ (x3 − cx) = 8 − 2c.
x→2+
x→2
For continuity, both limits have to be equal. Thus, for continuity we need 4c + 4 = 8 − 2c. Solving this
equation for c we get c = 4/6 = 2/3. The answer is c = 2/3. The reason the function is continuous with this
choice of c is that now we have limx→2− f (x) = limx→2+ f (x) = f (2) = 20/3; that is, the limit of f (x) as x
approaches 2 exists and equals f (2).
COMMENT: This exercise asked you “To find the value of the constant c.” Now, you may not know what
the value of c is, you might not even understand 90% of the exercise. But one thing should be clear: The
only possibly correct answer must say what c is. Any answer that does not include a sentence that begins
with c = , or declaring that no value of c works, is certainly wrong. Generally speaking, you should answer
the questions asked in the exercise, not go off along some irrelevant tangent.
Among the tangentialists, some of you talked of two functions. There is only one function in this exercise.
2. Find the limit or show that it doesn’t exist. Show some work. Don’t go overboard. If you can figure out a
limit by inspection, just write the answer. Of course, if your answer is wrong, you don’t get partial credit if
I don’t see any work. Write out your answer in the form of a sentence. For example: “limx→3 x = 3”
is a sentence; “3” is not. (This exercise has parts (a),(b), (c), and (d).)
3x − 2
.
2x + 1
Solution. (Exercise 15 of §2.6) Since the expression (not the equation!) is a quotient of two polynomials of the same degree 1, the limit is the quotient of the leading coefficients:
3x − 2
3
lim
= .
x→∞ 2x + 1
2
(2x2 + 1)2
(b) (15 points) lim
.
x→∞ (x − 1)2 (x2 + x)
Solution. (Exercise 21 of §2.6) The expression is a quotient of two polynomials of degree 4. If we
multiply them out, the numerator is 4x4 + 4x2 + 1, the denominator is x4 − x3 − x2 + x. The quotient
of the leading coefficients is 4/1 = 4. Thus
(2x2 + 1)2
=4
lim
x→∞ (x − 1)2 (x2 + x)
√
(c) (15 points) limx→−∞ (x + x2 + 2x). Notice that the limit is to −∞, not to ∞.
Solution. (Exercise 26 of §2.6) The information that the limit was for x going to −∞ was an important
one. I wonder how many people wondered about it?
√
√
√
x − x2 + 2x
(x2 − (x2 + 2x)
2
2
√
√
lim (x + x + 2x) =
lim (x + x + 2x) ·
= lim
x→−∞
x→−∞
x − x2 + 2x x→−∞ x − x2 + 2x
−2x
√
=
lim
.
x→−∞ x −
x2 + 2x
(a) (15 points) lim
x→∞
2
Here is where a bit of thinking could help. x is going to minus infinity.
That means that x− anything
√
positive will also go to −∞. If one simply keeps in mind that x2 = |x| = −x, not x (for negative
x), one is on the correct path. One already begins to realize that the limit should be −1. Perhaps
the easiest way to see this is to use something I mentioned (very briefly, I admit) in class; namely that
limx→∞ f (x) = limx→−∞ f (−x). Then we get
−2x
√
x→−∞ x −
x2 + 2x
lim
2x
2x
x
√
√
= lim
x→∞ −x −
x2 − 2x x→∞ − xx − x2x−2x
2
2
2
√
√
√
= lim
= lim
= lim
2
2
x→∞
x→∞
x→∞
−1 − 1 −
−1 − x x−2x
−1 − xx2 − 2x
2
x2
=
lim
=
2
x
2
= −1.
−1 − 1
1 − ex
.
x→∞ 1 + 2ex
(Exercise 35 of §2.6)
(d) (15 points) lim
Solution.
1 − ex
lim
x→∞ 1 + 2ex
=
1−ex
x
lim e x
x→∞ 1+2e
ex
−1
1
e−x − 1
=
=−
x→∞ e−x + 2
2
2
= lim
In the next two exercises I need to see work. I need to see that you are using only concepts and ideas from
Section 2.7 (or earlier). In other words, don’t use anything you might have learned in a previous calculus
course (or elsewhere), but that hasn’t been seen yet in this course.
√
3. (15 points) Find an equation of the tangent line to the curve y = x at (1, 1).
Solution.
(Exercise 7 of §2.7)
√
The slope of the tangent line√is the derivative of the function of which y = x is the graph. That is the slope
will be f ′ (1) where f ′ (x) = x. Now
√
√
√
√
√
1+h− 1
1+h−1
1+h−1
1+h+1
′
f (1) = lim
= lim
= lim
·√
h→0
h→0
h→0
h
h
h
1+h+1
1+h−1
h
̸h
√
= lim √
= lim √
= lim
h→0 h( 1 + h + 1)
h→0 h( 1 + h + 1)
h→0 ̸ h( 1 + h + 1)
1
1
1
= lim √
=√
= .
h→0
2
1+0+1
1+h+1
Using now the point-slope form of the line, we get the equation in the form y − 1 =
1
1
1
(x − 1) or y = x + .
2
2
2
Either form is OK.
4. (10 points) Find f ′ (a) if f (x) = 3x2 − 4x + 1. (Remember: Show work!)
Solution.
f ′ (a)
(Exercise 27 of §2.7)
f (a + h) − f (a)
3(a + h)2 − 4(a + h) + 1 − (3a2 − 4a + 1)
= lim
h→0
h→0
h
h
3(a2 + 2ah + h2 ) − 4(a + h) + 1 − 3a2 + 4a − 1
̸ 3a2 + 6ah + 3h2 − ̸ 4a − 4h+ ̸ 1− ̸ 3a2 + ̸ 4a− ̸ 1
= lim
= lim
h→0
h→0
h
h
2
(6a + 3h − 4) ̸ h
6ah + 3h − 4h
= lim
= lim 6a + 3h − 4 = 6a − 4.
= lim
h→0
h→0
h→0
h
̸h
=
lim
The answer is f ′ (a) = 6a − 4.
3
Bonus Problem (10 points) By evaluating the function f (x) = x4 + x − 3 at x = 0, 1, 2, . . . find an interval of
length 1 in which the equation x4 + x − 3 = 0 must have a root. Explain.
Solution. We see that f (0) = −3, f (1) = −1, f (2) = 15. No need to go any further; the equation must have a
root in the interval [1, 2] because the function f is continuous, f (1) = −1 < 0, f (2) = 15 > 0; by the intermediate
value theorem there is c, 1 < c < 2 such that
0 = f (c) = c4 + c − 3.