IB HL Math Homework #4: Functions and Equations Solutions
Assigned: 9/20/07, Thursday
Due: 9/25/07, Tuesday
1) Find all of the values of x that satisfy the inequality
2x
 1.
| x 1|
Split the work into two cases, one for x > 1 and one for x < 1. (Note that x ≠ 1.)
2x
 1 , so 2x  x  1 , and x  1 . We can do the first step
x 1
because we know that in this case x – 1 is positive. our conclusion tells us that if x >
1, then a solution would be x < -1. Thus, there are no values of x > 1 for which the
given statement is true.
If x > 1, then we have
2x
 1 , so 2x  1  x , and x  1/ 3 . We can do the first step
 ( x  1)
since we know that –(x – 1) is positive. Thus, our total set of solutions is {x: x < 1/3}.
If x < 1, we have
2) The polynomial f(x) = x3 + 3x2 + ax + b leaves the same remainder when divided by (x
– 2) as when divided by (x + 1). Find the value of a.
Carrying out the two required synthetic divisions, we get the following:
2
-1
1
1
1
3
5
2
a
b
10+a 20+2a+b
a–2 2–a+b
The two remainders are the same, so we have:
20 + 2a + b = 2 – a + b
3a = -18
a = -6
3) The functions f and g are defined by f : x  e x , g : x  x  2. Calculate
(a) f 1 (3)  g 1 (3)
(b) ( f  g ) 1 (3).
(a) f 1 ( x)  ln x , and g 1 ( x)  x  2 . (These can be computed in the usual way.)
Thus, f 1 (3)  g 1 (3)  (ln 3)(3  2)  ln 3
(b) ( f  g ) 1 ( x)  g 1 ( f 1 ( x)). Thus, ( f  g ) 1 (3)  g 1 ( f 1 (3))  g 1 (ln 3)  ln 3  2 .
4) A sum of $5000 is invested at a compound interest rate of 6.3% per annum.
(a) Write down an expression for the value of the investment after n full years.
(b) What will be the value of the investment at the end of five years?
(c) The value of the investment will exceed $10000 after n full years.
(i) Write an inequality to represent this information.
(ii) Calculate the minimum value of n.
(a) ($5000)(1.063) n
(b) Value = ($5000)(1.063) 5  $6786.35.
(c) (i) ($5000)(1.063) n  $10000
(1.063) n  2
n ln( 1.063)  ln 2
ln 2
n
ln 1.063
n  11.3
Thus, the minimum number of years necessary to double the investment is 12 years.
5) Let f ( x) 
f ( x)  g ( x) .
x4
x2
, x  1 and g ( x) 
, x  4 . Find the set of values of x such that
x 1
x4
x4 x2

x 1 x  4
x4 x2

0
x 1 x  4
( x  4)( x  4)  ( x  2)( x  1)
0
( x  1)( x  4)
x 2  16  ( x 2  x  2)
0
( x  1)( x  4)
x  14
 0.
( x  1)( x  4)
The critical points are x = -1, 4, 14. The given statement is true when x < -1 and
when 4 < x ≤ 14. Thus, the solution is {x: x < -1 or 4 < x ≤ 14}
6) Given that (x – 1) and (x – 2) are factors of 6x4 + ax3 – 17x2 + bx – 4, find a and b,
and any remaining factors.
Use synthetic division:
1
6
6
a
a+6
-17
a-11
b
a-11+b
-4
a-15+b
So, a – 15 + b = 0
2
6
6
a+6 a-11 a-11+b
a+18 3a+25 7a+39+b
So, 7a + 39 + b = 0
Subtracting the first equation from the second, we get:
6a + 54 = 0, so a = -9. It follows that b = 24.
After dividing the original polynomial by x – 1 and x – 2, we have the following
quadratic: 6x2 + (a+18)x + (3a+25), which is 6x2 + 9x – 2. This does not factor
 9  129
"nicely." The other two roots are irrational. They are x 
. In factored
12
 9  129
 9  129
)( x 
).
form, we could write this as 6( x 
12
12
x
7) Consider the functions f ( x)  e 2 x and g ( x)  sin .
2
(a) Find the period of the function f  g.
(b) Find the intervals for which ( f  g ( x))  4.
(a) ( f  g ( x))  e 2 sin(x / 2) . The period of this function completely depends on the
periodicity of the exponent, since the exponential function is a monotonically
2
4.
increasing function. The period of g is
 2
(b) e 2 sin(x / 2)  4
 x 
2 sin    ln 4
 2
 x 
2 sin    2 ln 2
 2
 x 
sin    ln 2
 2
x
2
x
 sin 1 (ln 2) , within the first quadrant.
2 sin 1 (ln 2)

x  0.488 .
Since the sin function is decreasing in the second quadrant, and sin(x) = sin(π-x),
it follows that the other angle in the first or second quadrant at which
 x 
sin    ln 2 is 2 – 0.488 = 1.512. Also, since the period of the function is four, it
 2
follows that all solutions can be expressed as {x: 0.488+4k < x < 1.512+4k, k  Z }.
8) The function f is defined on the domain [-1,0] by f : x 
(a) Write done the range of f.
1
.
1 x2
(b) Find an expression for f-1(x).
(a) The desired range is [.5, 1]. The maximum occurs when x=0, which minimizes
the denominator, and the minimum occurs when x=1, which maximizes the
denominator.
(b) x 
1
1 ( f
1
( x)) 2
1
1  ( f 1 ( x)) 2 
x
1
( f 1 ( x)) 2   1
x
1
f 1 ( x)  
 1 , we take the negative because the domain of the original function
x
is negative.