Math 265
Simulated Midterm Exam 1
Solutions
Take note of how the points of each question are assigned.
1. = (2 pt)
= (1 pt)
(1 pts)
2. a) (2 pts)
Domain: 2x – 9 0; x 9/2. (1 pts)
b) (2 pts)
Domain: (1 pt)
3. Basic function (1 pt)
Shift to the left by 4 units (1 pt)
Vertical stretch by 3 units (1 pt)
(2 pts)
-4
4. a) = (1 pt)
The polynomial is continuous at 2. (2 pts)
b) = it does not exist (1 pt)
because
since ,
and (1 pt)
since ,
and (1 pt)
c) it does not exist (1 pt)
because
since
and (1 pt)
since
and (1 pt)
d) (1 pt)
(1 pt)
(1 pt)
e) = (2 pts)
= (1 pt)
5. a) By the power rule
(2 pts)
b) By the quotient rule
(2 pts)
c) By the chain and power rules
(2 pts)
d) By the General Power Rule
(2 pts)
6. Slope of the tangent line to is (1 pt)
Slope of the line is 6 (1 pt)
So we want x so that 6x + 4 = -1/6, (1 pt)
solving x = -25/36 (1 pt)
7. Volume of sand (1 pt)
since h = diameter = 2r the volume is (1 pt)
differentiating (1 pt)
hence for h = 4 and solving for
we have m/min. (2 pts)
8. (3 pts)
solving for y’
(2 pts)
9. = . (1 pt)
Let f(x) = , and , (1 pts)
hence
(1 pts)
(2 pts)
10. Answers vary. One point for each condition and one point for sketching an actual graph of a function.
-2
Sketches with overlapping lines, such as the one below, are not graphs of functions, since the vertical line test fails in this case.
-2