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Practice Final Solutions MATH 111 Winter 2004
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[11] Let f & g, be functions, and x & y be real numbers. T (f g)(x) = (gf )(x) Note f (x)g(x) = g(x)f (x). F (fg )(x) = ( gf )(x) Note sometimes f g^ ((xx)) 6 = g f( (xx)). F x^2 = y defines x as a function of y Note there are two values of x which satisfy x^2 = y when y > 0. T x − 2 is a factor of 12 x^4 − 2 x^2 + x − 2 Use the factor theorem and the fact that 12 (2^4 ) − 2(2^2 ) + 2 − 2 = 0. F log(log(e)) = 0. Note log(log(e)) ≈ − 0. 36221. T The diameter of a circle varies directly with the radius. Note d = 2r.
Right answers will not get credit without supporting work. Note ”unde-
fined” and ”no solution” are possible answers.
- [2] Define the absolute value of a real number c.
|c| =
c if c ≥ 0 , -c if c < 0.
- [2] Define log x = y
log x = y exactly when 10 y^ = x. (Alternatively, log x is the exponent you need to raise 10 to in order to get y.)
- [3] Find the equation for a line that is perpendicular to the line with end points (51, 60) and (53, 50).
The slope of the original line is m 1 = 6051 −−^5053 = (^) −^102 = −^15. Thus, our new line will have slope m 2 = 5 (since −^15 · 5 = − 1 ). Such a line is given by y = 5x.
- [3] Given kx^2 + 5x − 2 = 0, what does k have to be to ensure 2 real solutions? Give answer in interval notation.
We note that this polynomial has two real solutions when the descriminant 52 − 4 · k · − 2 > 0. Thus, 8 k > − 25 , so k > − 25 / 8. In interval notation, the solution is expressed by
−^258 , ∞
- [5] What is the diameter of the circle described by: 3x^2 + 9x + 30 + 3y^2 = 24y?
3 x^2 + 9x + 3y^2 − 24 y = − 30 x^2 + 3x + y^2 − 8 y = − 10 x^2 + 3x + 9/4 + y^2 − 8 y + 16 = 33/ 4 (x + 3/2)^2 + (y − 4)^2 = 33/ 4
Hence the radius is
√ 33 2 , and so the diameter is^
- [11] Given f (x) = (^2) x^1 +3 :
- [5] Compute the difference quotient. Recall the difference quotient is:
f (x) =
f (x + h) − f (x) h
1 2(x+h)+3 −^
1 2 x+ h
1 2 x+2h+3 −^
1 2 x+ h
=
h
2 x + 3 (2(x + h) + 3)(2x + 3)
2 x + 2h + 3 (2(x + h) + 3)(2x + 3)
2 x + 3 − 2 x − 2 h − 3 (2(x + h) + 3)(2x + 3)h
=
− 2 h (2(x + h) + 3)(2x + 3)h
(2(x + h) + 3)(2x + 3)
There is a vertical asymptote at x = −^32 , since this a root of the denominator and not the numerator. There are no holes. The denominator has a higher degree than the numerator, thus there is a horizontal asymptote at y = 0. There is no x-intercept (as the numerator has no roots); the y-intercept is at f (0) = 13. Since 2 x + 3 > 0 when x > 32 , sign analysis indicates that f (x) > 0 when x > −^32 , while f (x) < 0 when x < −^32.
- [2] Find the inverse graphically on the above graph.
Reflect over the line y = x.
(c) [4]
g f
(x)
g f
(x) = √^37 xx−−^73. We have to determine when 7 x − 3 > 0. This occurs on (3/ 7 , ∞), and thus that is the domain. (d) [4] f (g(x))
f (g(x)) =
7 g(x) − 3 =
7(3x − 7) − 3 =
21 x − 52. For the domain, we need g(x) to be in the domain of f , in other words, 3 x − 7 ≥ 3 / 7. So 3 x ≥ 52 / 7 , that is, x ≥ 52 / 21.
- [7] Given f (3) = 0, use the factor theorem to find the other roots of x^4 − 3 x^3 − 25 x^2 +75x
We know x − 3 is a factor of f by the factor theorem. Divide f by x − 3 , to get x^3 − 25 x. Note we can factor this to get x(x^2 − 25) = x(x + 5)(x − 5). Thus, the roots are 3 , 5 , − 5 , 0.
- [3] Simplify: 2 − log 5 (25z).
2 − log 5 (25z) = 2 − (log 5 (25) + log 5 (z)) = 2 − (2 + log 5 (z)) = − log 5 (z).
- [4] Solve for x: 4x^ − 3 ∗ 2 x^ = 10. hint: use a substitution
Note 4 = 2^2. So, we can write (2^2 )x^ − 3 · 2 x^ = 10, so (2x)^2 − 3 · 2 x^ = 10. Let u = 2x. Then we have u^2 − 3 u − 10 = 0, that is, (u − 5)(u + 2) = 0. So, u = 5 or u = − 2. So 2 x^ = 5 or 2 x^ = − 2. The second is impossible, so we have 2 x^ = 5, which means x = log 2 (5).
- [4] Draw the complete graph of f (x) = log 2 (^12 x + 3) + 1, using only graph transforma- tions. List the transformations in order.
The transformations are
- Horizontal shift left 3 units
- Horizontal shrink by a factor of 1 / 2
- Vertical shift up 1 unit
- [4] Your given a 16 oz mocha that is a rather weak 3% espresso. You, knowing you’ll be up late studying mathematics, would rather like a 30% espresso drink. Realizing this you purchase an espresso machine. How much weak mocha do you discard and replace with straight espresso?
Set up the following equation, where x is the amount of the drink to be discarded and replaced:
((16 − x) oz. of 3% espresso) + (x oz. of 100% espresso) = (16 oz. of 30% espresso).
This can be rewritten .03(16 − x) + x = 16(.3). We can solve for x to get x ≈ 4. 45361.
- [2] How long will a loan take to triple at 20% interest compounted quarterly?
The amount of money owed after t years is given by f (t) = P
) 4 t , where P is the original principal. We want f (t) = 3P , that is, 3 P = P (1 +. 2 /4)^4 t. We can immediately cancel the P ’s to get 3 = (1 +. 2 /4)^4 t, that is, 3 = 1. 054 t. Apply ln to both sides, obtaining ln(3) = ln(1. 054 t) ln(3) = 4t ln(1.05) ln(3) 4 ln(1.05)
= t
t ≈ 5. 62927
. So, you should leave your money in until third quarter of the fifth year.
- [2] After much research, Uncle Joe has found a function, P (x), describing his profit for the number, x, of wing-digs he produces. How many wing-digs should be make to maximize profit given that the function for profit is:
P (x) = −100(x − 75)^2 + 600
The graph of f is a parabola opening downward, and the maximum of f corresponds to the vertex of this parabola. You could multiply out to get the quadratic in standard form (ax^2 + bx + c), but it is easier to use the fact in the box on page 228, which states that if f (x) = a(x − h)^2 + k, then the vertex is at (h, k). Thus, Uncle Joe’s profit is maximized when he produces 75 wing-digs, and he will make a profit of 600.
- [10] Graph and answer the following for
f (x) =
(x − 2)(x^2 + 3x − 10) x^2 − 5 x + 6 .
(a) [3] Domain: (Give answers in interval notation) The domain comprises all real numbers that are not roots of the denominator. We can factor to get f (x) = (x−(2)(x−x3)(−2)(x−x2)+5). Thus, the domain is (−∞, 2) ∪ (2, 3) ∪ (3, ∞).
