Answers to practice problems. (Not complete solutions, but answers) On test you will be expected to show all work. For #5c) for example, you will be expected to show the completed table, filling in the columns for delta w, delta v, and delta v/delta w.
1. No, semester grade in Math 121 is NOT a function of grade on Test 1. You could make 90 on Test 1 and receive an A in the course, or a B in the course, or a C in the course, or a D in the course, or an F in the course, for example. There is a value of the independent variable, score on Test 1, for which there is more than one corresponding value of the dependent variable, semester grade.
2. Yes, cost of tuition at MTSU is a function of the number of credit hours taken. For each number of credit hours there is a certain tuition charged. (otherwise wouldn't there be chaos?)
3. P = 4s so s = P/4. But A = s^2. So s = sqrt(A). Thus P = 4 sqrt (A).
Final answer: P = f(s) = 4 sqrt(A) (Perimeter equals 4 times the non-negative square root of A)
4. C = f(x) = 9.72x
5. a) No w is not a function of v because if v = 10, then w = 4 or 5
b) Yes v = f(w). Given a value of w, there is just one corresponding value of v.
c) No, for b) slopes are not constant. To see this add columns to the table for delta v, delta w, and
delta v/delta w. The slope column is the delta v/delta w. The numbers in this column differ so the
function is not linear.
6. a) No. To see this add columns to the table for delta v, delta w, and
delta w/delta v. The slope column is the delta w/delta v. The numbers in this column differ so the
function is not linear.
b) No. To see this add columns to the table for delta v, delta w, and
delta w/delta v. The slope column is the delta v/delta w. The numbers in this column differ so the
function is not linear.
7. a) Yes To see this add columns to the table for delta r, delta t, and
delta t/delta r. The slope column is the delta t/delta r. The numbers in this column
are the same, specifically 2, so the function is linear with a slope of 2. The symbolic representation
t = f(r) = 2r-2
B) Yes, delta r/delta t is constant, namely 1/2, so that function is linear with a slope of 1/2 The
symbolic representation is r = f(t) = (1/2)t + 1
8. a) q = f(z) = 1 (This is a constant function. Graph is a horizontal line)
b) No z is not a function of q. If q = 1, z could be 1 or 2 or 3 or 4
9. a) y = 2x-1
b) Oops, poorly worded question. If a line has slope 0 and x-intercept 4, it has to be a horizontal line through the point (4,0). So the line is the x-axis which has the equation of y = 0 (or 0x + y = 0). The reason I say this is a poorly worded question is that there are infinitely many x-intercepts, not just one.
c) x = 1 (or x + 0y = 1, if you prefer)
d) y = (2/3)x + (7/3)
e) y = (-2/3)x + 1
f) y = (3/2)x - (13/2)
10. f(4) = -2; f(3) = 1
11. a) 3 b) 1 c) 1 d) 3 e) There are 4 solutions. They are: x = 1, x is approx 2.7, x is approx 2.4, x is approx -2.5
12. a) 0
b) (sqrt(3))/4
c) (sqrt(a-1))/a
d) (sqrt(a))/(a+1)
e) (sqrt(x))((x+1)
13. a) No, there is a value of t for which there is more than one corresponding value of v. t = 0 for example
b) Yes, for each value of v there is just one corresponding value of t. (Graph passes vertical line test. We can use VLT for b) since the IV v is graphed on the horizontal axis)
14. Inversely proportional. As the number of people sharing the house increases, the cost per person decreases. C = k/N where N is the number of people sharing the house and C is the cost per person.
To find k:
| N | C |
| 1 | 1350 (150*9) |
1350 = k/1
k = 1350
C = 1350/N
If C is $100, N = 1350/100 so N = 13.50. For each person to pay a max. of $100, you would have to have 14 people.
15. 4
16. 5/3
17. z = 3t
18. Oops, typo. We don't need the dollar sign, do we? P = f(t) = 500 + 20 t
19. 7/3