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) . x4 x2 , x 1 and g ( x) , x 4 . Find the set of values of x such that x 1 x4 x4 x2 x 1 x 4 x4 x2 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.