Math 131 Exam 1 Practice Answers
Tom Linton,
Central College, Spring 2000
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The period is about 3.14 (actually its Pi).
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The graph of a function f(x) is shown below.
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f(x) is decreasing from x = 0.3 to x = 3.3.
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f(3) is about -5 and f(x) = 0 for x = 1.5 and x = 4.4.
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f(x) is concave up from x = 2.5 to x = 5. The inflection point of f(x)
is at about (2.5, -2.5).
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f(x) has a valley bottom (local minimum) at x = 3.3.
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Shown below is a plot of the temperature (in degrees Fahrenheit) of a
warm
beverage which was put in a cold refrigerator. The input is time
measured
in minutes, with t = 0 corresponding to the time when the beverage was
put in the refrigerator. One (exact) point on the graph is t = 2, Temp
= 75 and the formula for the beverage's temperature has the form B(t) =
c + a* bt (a, b, c are constants, t is the variable).
Use the plot to answer the following questions.
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c equals about 32.
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f(2) / f(0) = (a*b2) / (a) = b^2. (75-c) / (80-c) = f(2) /
f(0)
because (2, 75 - c) and (0, 80 - c) are on the graph of f(x). f(0) =
a*b^0
= a. Using c = 32, I get b^2 = 43 / 48, so b = 0.9465. Since a = f(0) =
48, we have a = 48.
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B(t) = 50 for t = about 17.84 (but this depends on the values you chose
for a, b and c).
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Solving c + a*bt = 50 for t, yields t = ln( (50 - c) / a) /
ln(b) in the general case, or numerically (for my values of a,b and c)
t = 17838.
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B-1(50)
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B-1(40) is about 32.587
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An algebraic representation of the function f(x) and a plot of the
function
g(x) are given below. Use these representations of the functions to
answer
the following questions.
| f(x) = |
4,
4 - 3x, |
if x <= 0
if x > 0 |
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f(1) = 1and g(1) = -7.
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f(2/3) = 2.
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g(2) = -20.
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Let h(x) = g( f(x) ). h(1/3) = g( f(1/3) ) = g(3) = 22.
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x = 2/3 has h(x) = -20. You need g(2) and f(2/3) = 2.
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A graph of 4x3 - 1.9x for 0 < x < 25,
has two zeros at x = 0.7377 and x = 14.733 and a hilltop at (12.4098,
4765.36)
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The graph below has the form f(x) = A*sin (Bx + C) + D, where A = 8, B
= 3.14, C = 5.97 and D = 5.