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September 28, 2004 Math 111 C — Exam I
Show all work clearly for partial credit. Do not use the graphing capabilities of your calculator.
- (8 points) Find exactly; i.e., do not use a calculator:
(a) log 2 10 + 2 log 2 6 − log 2 45 (b)
ln 5
ln 5
- (20 points) Find the limits, if they exist:
(a) lim x→ 1
x^2 − 3 x + 2 x^2 − 1
(b) lim x→ 2
x^2 + 3x + 2 x^2 − 1
(c) lim x→ 0
ln(x^2 ) (d) lim x→ 0 −
( 1 |x|
x
)
- (14 points) For the function f (x) with the graph below, find or approximate (if they exist):
(a) f (−2), (b) limx→− 2 f (x), (c) the equation(s) of the vertical asymptote(s), (d) the equation(s) of the horizontal asymptote(s), (e) limx→ 0 f (x), (f) the x-value(s) at which f has a removable discontinuity, and (g) the slope of the tangent line to the graph of y = f (x) at x = 1.
- (18 points) Find the equations of the horizontal and vertical asymptotes:
(a) f (x) =
x^2 − 4 x^2 − 3 x + 2
(b) g(x) = √^2 x x^2 + 4
- (17 points) (a) Write down in terms of h an expression for the slope of the (secant) line joining the points on the graph of the function f (x) =
2 x + 3 with the x-values x = 11 and x = 11 + h. (b) Using your answer to (a), find the slope of the tangent line to the graph of the function f (x) =
2 x + 3 at the point x = 11. (Use of a derivative formula — whatever that is — or simply the answer will receive no credit. Note: This is the same as the instantaneous rate of change of f (x) =
2 x + 3 with respect to x at the point where x = 11.) (c) Using your answer to (b), write the equation of the tangent line to the graph of√ f (x) = 2 x + 3 at x = 11.
- (15 points) Let f (x) =
x, a = 9 and ε = 0.4. What is the largest value of δ for which,
if 0 < |x − 9 | < δ , then |
x − 3 | < .4?
(An answer of the form “the smaller of the two numbers and ” is preferred but not required. The following drawing may be helpful.)
- (8 points) The area of an algae growth on the surface of the liquid in a vat doubles every 10 hours. At midnight the area was 4 cm^2. Write a formula in terms of t for the area A(t) of the algae growth t hours after midnight.
Some possibly useful equations:
y − y 0 = m(x − x 0 ) a^3 − b^3 = (a − b)(a^2 + ab + b^2 )
ar^ = b ⇐⇒ loga b = r loga c =
logb c logb a
(c) Because f (11) = 5, the desired tangent line is y − 5 = 15 (x − 11).
- Because the inverse of the function y =
x is x = y^2 , the question marks are 2. 62 and 3. 42 ; so the answer is: the largest δ that works is the smaller of the two numbers 9 − 2. 62 [= 2.24] and 3. 42 − 9[= 2.56], [i.e., 2.24]. It would not have been necessary to give any of the material in the brackets.
- This is an example of “exponential growth”, so the formula is A(t) = A 0 (2kt), where A 0 is the area at time 0 (which we are told is 4) and the “growth rate” k is a constant chosen to fit the given information. (We could have used a different base in place of 2, and then k would have been different.) In this case, because we know that, when t = 10, A(10) is twice as large as it was at midnight, so it is 8: 8 = 4(2k(10)), so 1 = k(10), so k = 1/10. So the formula is A(t) = 4(2t/^10 ).
