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Math 111C Exam I Solutions, Exams of Calculus

Solutions to exam i of math 111c, including calculations for logarithmic and limit problems, finding equations of asymptotes, and determining the slope of tangent lines. It covers topics such as logarithms, limits, functions, and calculus.

Typology: Exams

Pre 2010

Uploaded on 08/19/2009

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koofers-user-f6y 🇺🇸

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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.
1. (8 points) Find exactly; i.e., do not use a calculator:
(a) log210 + 2 log26log245 (b) ln 5
125
ln 5
2. (20 points) Find the limits, if they exist:
(a) lim
x1
x23x+ 2
x21(b) lim
x2
x2+ 3x+ 2
x21
(c) lim
x0ln(x2) (d) lim
x01
|x|+1
x
3. (14 points) For the function f(x) with the graph below, find or approximate (if they exist):
(a) f(2),
(b) limx→−2f(x),
(c) the equation(s) of the vertical asymptote(s),
(d) the equation(s) of the horizontal asymptote(s),
(e) limx0f(x),
(f) the x-value(s) at which fhas a removable discontinuity, and
(g) the slope of the tangent line to the graph of y=f(x) at x= 1.
4. (18 points) Find the equations of the horizontal and vertical asymptotes:
(a) f(x) = x24
x23x+ 2 (b) g(x) = 2x
x2+ 4
pf3
pf4

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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.

  1. (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

  1. (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

)

  1. (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.

  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

  1. (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.

  1. (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.)

  1. (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).

  1. 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.

  1. 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 ).