| 1.
y" changes sign at x =-5 so there is a point of inflection at x = -5. |
Answer D |
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2.
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Answer B |
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| 3.
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Answer C |
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| 4.
A is a restatement of the Mean Value Theorem for Derivatives. So it
is true all the time with the given conditions.
B is not true necessarily unless f(a)=f(b).
C is true because f is continuous on the closed interval and
differentiable on the open interval.
D is true because f is continuous on the closed interval and
differentiable on the open interval.
E is true since f is continuous on the closed interval. |
Answer B |
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5.
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Answer E |
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| 6.
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Answer A |
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| 7.
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Answer E |
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8. If h(x)=f(x)g(x) then
 usually, but we are
told that  so either g(x)=0
or f '(x)=0. We are told that g(x) >0 for all x, so then f '(x) must
be zero. This leads to f being a constant. Since f(0)=1, then
f(x)=1 |
Answer E |
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9. The total number of barrels of oil that passed
through the pipeline is 
or the area bounded by the function, t=0, t=24, and the t axis. This
approximately 600+600+600+600+600 or 3000 barrels. |
Answer D |
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| 10.
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Answer D |
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| 11. Since f is a linear function, f ' (x) is a
constant, and f "(x) is zero.
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Answer A |
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| 12.
Therefore the limit from the left does not equal the limit from the
right, no limit exists at x=2. |
Answer E |
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| 13. f is not differentiable at x=0 because of the
discontinuity or x=2 because of the vertical tangent line. |
Answer B |
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| 14.
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Answer C |
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| 15.
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Answer D |
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| 16.
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Answer E |
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| 17. From the graph we know that f(1)=0, f '(1)>0, and
f "(1) <0 so the correct order is f "(1) < f(1) < f '(1) |
Answer D |
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18.
 Tangent line:
y=1(x-0)+1 or y = x+1 |
Answer B |
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19. Since the second derivative is given, we know the
second derivative equals zero at x=0, x=-1, and x=2. The sign study of
f " is
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(-1,0) |
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| positive |
positive |
negative |
positive |
So there are two points of inflections. One at x=-1 and one at x=0 |
Answer C
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20.
 can be thought of as
finding the area under the function
 starting at x=-3.
Starting at -3 and finding area to the right of -3 will always yield a
positive answer. Starting at -3 and finding area to the left of -3
will yield a negative answer. Starting at -3 and integrating to -3 is
the only place that will yield zero area.
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Answer A |
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| 21.
\
so the only answer of this form is B |
Answer B |
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| 22.
The sign of f ' only changes at x=0. When x >0 f ' >0.
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Answer C |
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| 23. From the given graph of f we know that just to the
right of x=0 (call this point D) the slope is zero. As x approaches
positive infinity and negative infinity the slope also approaches zero.
Two graphs are possible so far. A and C. Studying the slope to
the right and left of the location D we notice the slope to the right is
negative, to the left it is positive. This rules out graph C. |
Answer A |
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| 24. First find a(t) by differentiating v(t).
Differentiating a(t) to see how a(t) is changing:
a(0)=12
a(3)=27-18+12=21
We can use the closed interval test. Checking the values of a(t) at
the critical values and the endpoints. From this we can see that the
maximum acceleration is at t=3 or 21. |
Answer D |
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25.
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Answer D |
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| 26. Setting up a sketch of this problem might help:
The two given points have been graphed. The red line is f(x)=1/2.
The red dot is the only value of k that provides two solutions. If
k=1/2 there would be one solution. If k = 1, 2, or 3 there would no
solutions.
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Answer A |
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27.
 By u-substitution:
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Answer A |
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| 28.
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Answer E |
