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(math005)[2002](f)mid~PPSpider^sol_10387.pdf
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Solutions to Midterm 1 (White: Version A), Math 005 Fall 2002
Part I: 10 multiple choice questions.
1. If f(x)= g(x + 3) and g(x)= x2 +2x + 1, compute f(1). Observe g(x)=(x + 1)2 .
(b) f(1) = g(1+3) = g(4) = (4+1)2 =25 .
2. Jenny has decided to pay o. two loans on which she has not made any previous repayment. The .rst loan is a loan in which she obtained $9, 000 two years ago at an annual interest rate of 8% compounded semiannually. The second is a loan where $12, 500 is due two years from now at an annual interest rate of 16% compounded quarterly. To the nearest dollar, how much is due now?
0.08 0.16
).8
(d) $9000(1 + )4 + $12, 500(1 + = $19, 662 .
24
x2 . 1
3. The domain of the function de.ned by the formula f(x) = is:
2x3
since division by 0 is unde.ned.
(a) All real numbers except 0
x2 +1 1
5
2
52
2x .3 .3
(c) f.(x)= . ()x = x +.
22 2x
4. The derivative of f(x)= is:
.
2 x
3
2
2 . 2x +1 2 . 2x +1 (x . 1)(x . 1) (x . 1)
xx
5. Find the limit lim .
(d) 0
since lim lim lim= = . x2 +2x . 3 x2 +2x . 3 (x . 1)(x + 3) (x + 3)
x1 x1 x1 x1
6. Determine the tangent line to the graph of the function y = f(x)= .6x2 +3x . 2 at the point (2,f(2)).
(a) y = .21x + 22 since f(2) = .20,f.(2) = .21 and the tangent line to the graph at (2,f(2)) is given
by y = f(2) + [f.(2)](x . 2) = .20 . 21(x . 2) = .21x + 22.
2 . 16).7. Compute g.(4) for g(x) = (31 . 15
x)(x
(e) 8.
Use the product rule (and the power rule) to
.15
compute g .(x)=( )(x 2 . 16) + (31 . 15 x)(2x), whence g .(4) = 0+(31 . 30)(8) = 8.
2x
8. Conversion between temperature measured in Fahrenheit F and measured in Celsius C is a linear function. Water freezes at 32F and 0C, and boils at 212F and 100C. The temperature in New York City is 68F. What
is this temperature in C?
5
(a) 0 + (68 . 32)=20 .
9
Required slope is 100 . 0 100 5
= =.
212 . 32 180 9
y = .49
(a) , min.
8
9. Find the maximum/minimum of the function y = f(x)=2x2 +5x . 3. since the
.49
parabolic graph opens upwards (why?) and its vertex is at ( .5 , ).
48
10. A vertical pole of height 4 metres snaps at a height of x metres above the ground. The upright por-tion (of x-metres) now forms the perpendicular side of a right-angled triangle, the snapped upper portion forms the hypoteneuse of the right-angled triangle (with the base of the triangle being along the ground). Determine the function f(x) that gives the area (in square metres) of the relevant right-angled triangle.
x
(c) f(x)= 16 . 8x .
2 We have a right-angled triangle with the length of its perpendicular = x metres
and the length of its hypoteneuse =4 . x metres. Thus, using the Pythagorean Theorem, the base has
length = .(4 . x)2 . x2 = 16 . 8x. The area is now given by half height times base.
Part II: 4 long questions.
11. For both parts (a) and (b), assume that the interest rate is 8% per year compounded every 6 months. Show your calcul