Solutions to Test #1 –MATH 1130 (#1) Assume x is positive. Simplify the radicals. p
p 12x2 + 3 3 x
p
p p p p p p 4 3 x2 + 3 3 x 36 2 p p p = 2 3 x+ 3 3 x 6 2 p p = 5 3 x 6 2:
72 =
(#2) Here’s a third-degree polynomial: f (x) = 3x3 + x2
27x
9:
(a) How many complex (real and nonreal) zeros must f have? 3. It’s a third-degree polynomial with real coe¢ cients. (b) Factor by groups and …nd ALL of its zeros. 3x3 + x2
9 = x2 (3x + 1)
27x
2
9 (3x + 1) = 0
(x + 3) (x
3) (3x + 1) = 0
x
x = h)2 + (y
(#3) The standard form for a circle is (x
3; 3;
9 (3x + 1)
1=3:
k)2 = R2 : Complete the square twice.
Find the location of the center of the circle and its radius: x2
2x + ( 1)2 + y 2 + 4y + 22 (x
The center is located at C (1; (#4) Let y = f (x) =
1)2 + (y + 2)2 = 25 = 52 :
2) and the radius is R = 5:
1 : Find the formula for f x+1 x =
= 20 + 1 + 4
1 (x) :
Exchange y for x: Solve for y:
1 ) xy + x = 1 ) xy = 1 y+1
y = f
1
(x) =
1
x x
=
1 x
x
1:
(#5) Symmetry (no work necessary): The answer to each of these clues is exactly ONE of these: (i) graph has y-axis symmetry (ii) graph is symmetric with respect to the origin (iii) neither. (ii) A graph of an “odd” function like f (x) = x5 has this. (i) A graph of an “even” function like f (x) = x4 has this. (i) A graph which satis…es f ( x) = f (x) has this. (ii) A graph which satis…es f ( x) =
f (x) has this. 1
(iii) The graph of f (x) = x3 + x2 has this. f ( x) = ( x)3 + ( x)2 =
x3 + x2 6= x3 + x2 or
x3 + x2 :
(#6) This is a THIRD-degree polynomial. Look at the graph of y = f (x) : The clue said to solve: 3x2 + 4x
4 = (x + 2) (3x x =
2) = 0
2; 2=3:
We already know that there is a relative maximum at x =
2: Thus, we must want x = 2=3:
(a) Find the open intervals where f is DECREASING: ( 2; 2=3) : (b) Find the open intervals where f is INCREASING: ( 1; (c) What is the multiplicity of the zero at x = does not cross.
2) ; (2=3; +1) :
2? 2. The curve touches the x-axis, but
(#7) This style of window features a semicircular area on top of a rectangular area. In this case, the two areas are combined into one piece of glass. Thus, the metal stripping perimeter consists only of the two vertical segments (y) ; the lower horizontal segment (x), and the semicircular piece on top (x=2) : The dotted line is NOT a piece of metal (just there for reference).
Let x = the length of a horizontal segment (bottom). Let y = the length of a vertical segment (there are 2).
(a) Find an expression for the total amount (in feet) of metal stripping necessary to hold the window together. Set this equal to 16 feet. x+
2
x + 2y = 16
(b) The area of the rectangular portion is A = xy: Solve the equation from part (a) for y and substitute this into A: This should give you a quadratic function A (x) : 2y = 16
1+
A = xy = x 8
2
x)y=8 1 1+ x 2 2
1 1+ x 2 2 =
1 1+ x2 + 8x = A (x) : 2 2
This is a concave downward parabola. (c) Find the value of x which MAXIMIZES A (x) : Simplify your answer as much as possible. x=
b = 2a
8 2
1 1+ 2 2
[This also gives us y = 4:] 2
=
8 1+
= 2
16 : 2+
(#8) Let f (x) = x1=3 : Let g (x) be the new function after the following transformations: Shift the curve to the RIGHT by 5 units and then DOWN by 2 units. (a) Give the formula for g (x) in terms of f (x) : g (x) = f (x
5)
2:
(b) Write out the formula for g (x) in terms of x: Leave the rational exponents in. 5)1=3
g (x) = (x (#9) The Point-slope Form in Larson is y
y1 = m (x
A line passes through P1 (1=3; 5) and P2 (1;
2:
x1 ) :
1) :
(a) Give the Point-slope Form as stated above. m=
1 1
5 6 = = 1=3 (2=3)
6
3 2
=
9:
We can use either point as our base point. y
5=
1 3
9 x
or y + 1 =
9 (x
1)
(b) Expand the expressions and then give the Slope-intercept form: y = mx + b: They are both equal to y = 9x + 8: (#10) Assume that h 6= 0: Simplify the di¤erence quotient: f (x + h) h
f (x)
; when f (x) = 2x2
x:
Substitute, expand, and then simplify. 2 (x + h)2
2x2
(x + h) h
x
= =
Since h 6= 0; this is a safe cancel.
f (x + h) h
[The limit when h ! 0 is 4x
f (x)
2x2 + 4hx + 2h2
x
h
2x2 + x
h 4hx + 2h2 h
h
= 4x + 2h
1:
=
h (4x + 2h h
1)
1:]
(#11) Power functions are always of the form: f (x) = kxp ; where p is a real NONZERO number (could be negative). Suppose we have the composition: h (x) = (f
g) (x) =
1 : ( 5x + 1)3
Find f (x) where f is a POWER FUNCTION and g (x) is whatever makes sense after you choose f: OUTER: f (x) =
1 =x x3
INNER: g (x) =
5x + 1:
3:
3
(#12) Use long division to divide. Write your …nal answer in the form: q (x) +
r (x) ; where q (x) is the quotient and r (x) is the remainder. x2 + 1
Divide this:
x4
2x2 + 3x x2 + 1
1
=???
I just wrote out the coe¢ cients.
1 0 1 )
The quotient is 1x2 + 0x
1 1
0 0
1
0
3
2 1
3
1
3 3
3 0
1 3
3
2
3 and the remainder is 3x + 2: x4
2x2 + 3x x2 + 1
1
= x2
3+
3x + 2 : x2 + 1
(#13) Given the piece-wise function: h (x) =
x + 1; x2 3;
x 1 x>1
(a) Evaluate h (1) : Use the …rst de…nition. h (1) = (1) + 1 = 0: (b) Evaluate h (2) : Use the second de…nition. h (2) = (2)2 3 = 1: (c) I did half of the graphing work. The open circle is located at (1; parabola would have started, but we’re not allowed to have x = 1: y
4
3
2
1 0 -4
-3
-2
-1
0
1
2
3
4 x
-1
-2
-3
-4
4
2) ; where the
(#15) Suppose we have the rational function: f (x) =
x2 x 6 x2 1
(a) Rewrite f in FACTORED form (numerator and denominator). f (x) =
(x + 2) (x (x + 1) (x
3) 1)
(b) Write the equations for the VERTICAL ASYMPTOTES. The denominator equals zero when x = 1 and the numerator is not zero. Thus, these vertical lines are the horizontal asymptotes. (c) Write the equation for the HORIZONTAL ASYMPTOTE. EXPLAIN how you are executing the appropriate rule. Since the degree of the numerator is the same as the degreee of the denominator (both equal to one), then the horizontal asymptote is the ratio of the leading terms. y= (#16) Here’s a parabola: y = g (x) = a (x
1 x2 = 1: 1 (x2 )
h)2 + k:
The vertex is located at V ( 2; 3) = (h; k) : This gives us y = a (x + 2)2 + 3: The other point is located at (0; 1) ; so we substitute and solve for a: 1 = a (0 + 2)2 + 3 ) 4a = The standard form is g (x) =
2)a=
1 : 2
1 (x + 2)2 + 3: 2
(#17) The average rate of change (AROC) is the slope of the line segment which connects the two points. p Find the AROC in f from x1 = 1 to x2 = 6 for the function f (x) = 1 x + 3: AROC =
f (6) 6
f (1) ( 2) ( 1) = = 1 5
1 : 5
(#18) The coe¢ cients of a …fth-degree polynomial f (x) are all integers. It turns out that x =
1 + 3i is one of the nonreal zeros.
(a) Suppose we know that there is only one other nonreal zero. Name it! It’s the complex conjugate: 1 3i:
5
(b) Assuming that you have correctly answered part (a), then that means there is a QUADRATIC (2nd degree) factor which divide into f (x) which will leave a zero remainder. Show how to …nd that factor ax2 + bx + c : x = x+1 = 2
1
3i
3i
= 9i2 =
(x + 1) 2
x + 2x + 1 =
9
9 ) x2 + 2x + 10 = 0:
The quadratic factor must be x2 + 2x + 10 : (c) Assuming that there are only two nonreal zeros, then how many real zeros must there be? You only get 2 points in you can give the proper name of the cool theorem... The FUNDAMENTAL THEOREM OF ALGEBRA says that there must be exactly 5 complex zeros. If we assume that the multiplicities were one, then there must be 5 2 = 3 real zeros. If you thought possibly that the multiplicities were two on both nonreal zeros, then the answer is 5 4 = 1 real zero. Either answer is acceptable, but you MUST explain your assumptions. (#19) This third-degree polynomial has one rational zero: x = zeros.
1=2 and then two more (trickier?)
We suggest that you try synthetic division. Find the other two zeros. f (x) = 2x3
3x2 + 2x + 2 =
x+
1 2
(???) = 0:
(#20) Suppose the center of a circle is located at (0; 0) and a radius is drawn between the center and the point (4; 3) : (a) Dead Greek guy. Find the length of the radius of the circle (distance). q p (4 0)2 + (3 0)2 = 25 = 5: (b) What is the slope of the line segment from (0; 0) to (4; 3) [solid line segment]? y 3 = x 4
0 3 = = m: 0 4
(c) What is the slope of the line segment (dashed) which is perpendicular to the radius? m? =
