Selected Solutions for Sample Final
8. h(x)=f(g(x)) so h'(x)=f'(g(x)) g'(x)
The g and g' functions are on the "inside". The only values of
g we know are when x=2. Thus A and B can be ruled out, because
x=1 in those.
h(2)=f(g(2))=f(1)=2
h'(2)=f'(g(2)) g'(2) = f'(1) 6 = (3)(6) = 18
Thus only C is correct.
11. Exponential decay decays by the same proportion (or percentage)
at each point in time. In our case
2.10/3.00 = 0.7 = 70%
Thus (6.10)(0.7) = 4.270 and so the answer is D.
12. If a graph is concave up then f''>0. This means f' is increasing.
f' is the slope of the tangent line, so C is correct.
17. Make the substitution
2
u = x + 4
Then
du
---- = 2 x
dx
It follows that
/ /
| x |
| ------ dx = 1/2 | 1/u du
| 2 |
/ x + 4 /
Therefore the answer is C.
18. The formula for future value of an income stream is
M
/
|
| S(t) exp(r(M - t)) dt
|
/
0
To find the answer we set S(t)=A a constant to solve for,
M=5 and r=0.07. Thus
5
/
|
50000 = | A exp(0.35 - 0.07 t) dt
|
/
0
and we find that
50000
A = -------------------------- = 8351.875520 $/year
5
/
|
| exp(0.35 - 0.07 t) dt
|
/
0
Therefore B is the correct answer.
19A. The line passing through the point (q,p)=(300,4) with
slope (-10/.5) is given by
q - 300 = -(10/.5)(p-4)
Thus q=380-20p.
19B. R = pq = p(380-20p) = 380p-20 p^2
19C. R'(p) = 380-40p
19D. Critial points when R'(p)=0. Thus critial when 380-40p=0
or when p=19/2.
19E. Since R''(p)=-40 < 0 then this is a maximum by the second
derivative test. Thus the price that maximizes revenue
is p=19/2 = $9.50.
20A. For part A you are approximating the integral
6
/
|
| r(t) dt
|
/
0
using a left hand sum with three terms. Since the interval [0,6]
is of length 6 then the width of each rectangle is 2.
A_left = r(0)*2+r(2)*2+r(4)*2 = 90.62741699