Pre-Calculus I - Answers

! Math 102 $ ANSWERS 1. x x Sketch the graph of f(x) = 1+ e . Label the intercept and asymptote. Sketch the graph of g(x) = ln (x–1) . Label the intercept and asymptote. f(x) = 1+ e x x The graph of f(x) = 1+ e is the graph of y = e shifted up 1 unit. x y = e has a horizontal asymptote at the x-axis, y = 0. x f(x) = 1+ e has a horizontal asymptote at y = 1. y=1 g(x) = ln (x–1) These functions are reflections of each other across the line y=x ; this is evidence of the fact that these two functions are inverses of each other. 2. Solve for x: ln (2x+1) + ln (x) = 0 ln (x (2x+1) ) = 0 ln (x (2x+1) ) = 0 e (x (2x+1) ) = 1 2x2 + x ! 1 = 0 (2x! 1)(x+1) = 0 x=1 To solve a logarithmic equation, we use exponential function. But first it would help to get one ln expression. Now we are ready to take the exponential function on each side e and simplify.... x = ½ or ! 1 x = ½ 5 x— 3 = 2 x . 3. Solve for x: ln 5 x— 3 = ln 2 x (x ! 3) ln 5 x ln 5 ! x ln 2 = 3 ln 5 x (ln 5 ! ln 2) = 3 ln 5 = And now, you MUST check these “solutions”. Since ln (! 1) does not exist, ! 1 does not solve the equation. ½ does: ln (2) + ln (½) = ln (2) ! ln (2) = 0 To solve an exponential equation, we use a logarithm. Taking ln of both sides, we get... Using the logarithmic property ln x p = p ln x ... = x ln 2 x ln 5 ! 3 ln 5 = x ln 2 x Resulting in a quadratic equation which we may solve by factoring, or by quadratic formula 3 ln 5 )))))))))) ln 5 ! ln2 Keep in mind that you are solving for x ! Math 102 $ ANSWERS 4. 1 Find the inverse of the function: f(x) = ))))) . 3x — 1 1 y = )))))) 3x ! 1 x = )))))) 3y ! 1 1 First we interchange x & y then we will solve for y.... 3xy ! x = 1 3xy = y = (10) starting by multiplying both sides by (3y ! 1) 1+x 1+x )))))) 3x f)1 (x) = 1+x )))))) 3x 5. Simplify as completely as possible: log2 8 – log2 (3) + log2 (6) + log2 (¼) + ln e 2 Several approaches are possible. But, since we don’t know the value of log2 3 or log2 6, it is log2 8 + log2 (6) ! log2 (3) + log2 (¼) + ln e 2 log2 2 3 + log2 (6' 3) + log2 (2 !1 ) + ln e 2 log2 2 3 + log2 ( 2 ) + log2 (2 !2 ) + ln e 2 3 + (12) 1 + !2 + 2 = 4 6. A population of beetles had 4 members on Monday, and 400 on Wednesday. How many will there be on Friday? 40,000 When will the population reach 100000? Intuitively: the population must be multiplying by 10 each day. That is: Mon 4 Tues 40 Wed 400 Thurs 4000 Fri 40000 et cetera.... t To answer the second question, we solve: 4C10 = 100000... rt Using “SOP”: A(t) = A0 e r2 400 = 4 e 2r 100 = e ln 100 = 2r r = (½) ln 100 = ln 10 SO A(t) = 4 e 4e e tln 10 tln 10 t ln 10 t tln 10 = 100000 = 25000 = ln 25000 = ln 25000 ln 10 )))))))) t 10 = 25000... t = ln 25000/ln 10. ! Math 102 $ ANSWERS 7. According to the rational zeroes theorem, only certain rational numbers may be zeroes of the polynomial 2 3 function: P(x) = 8x — 36x — 2x + 9. List all the theoretically possible rational zeroes of P(x). Rational Zeroes Theorem says the only rational possible zeroes are p/q where p divides 9, q a factor of 8. This gives us: 9, 3, 1 ± )))))))) 8, 4, 2, 1 = 1 3 9 1 3 9 1 3 9 ± 1, 3, 9, ), ), ), ), ), ), ), ), ) 2 22 4 4 4 8 8 8 8. Use the process of synthetic division to test the numbers ½ and —½ for possible zeroes of the polynomial 2 3 function P(x) = 2x — 5x + 5x + 4. ½ 2 !5 2 1 !4 5 !2 3 !½ 4 3/2 5½ 2 2 !5 !1 !6 5 3 8 4 !4 0 ... telling us that P(½) = 5 ½ and P(! ½) = 0 So ! ½ is a zero of P. And P(x) = (x+ ½) (2x2 ! 6x + 8 ) = (x+ ½ )2 (x2 ! 3x + 4 ) 9. Now finish the job: Find ALL the zeroes of the polynomial function given in #8. The remaining two zeroes of this polynomial function can be found in Q(x) = x2 ! 3x + 4 . 3 ± r%%!%%7 Using the quadratic formula, Q(x) = 0 when x = ))))))) 2 3 2 The zeroes of P(x) = 2x — 5x + 5x + 4 are !1'2 and 3'2 ! '2 and 3'2 + !7 r%% % !7 '2 r%% % 10. Write a polynomial function with real coefficients having zeroes: 1 a zero of multiplicity two, and i a zero. The value of the polynomial at x=2 must be 15. 1 a zero of multiplicity two ¸ (x! 1) 2 is a factor of P. i a zero, and coefficients real ¸ ! i is also a zero, and (x ! i) and (x + i) are factors of P. Thus: 2 P(x) = A (x! 1) (x ! i) (x + i) = A (x! 1) 2 (x2 + 1) And P(2) = A (2!1) 2 (22 + 1) = 5A ...which must be 15, so A is 3. So P(x) = 3 (x! 1) 2 (x2 + 1) P(x) = 3x4 ! 6x3 + 6x2 ! 6x + 3 is a polynomial function satisfying all the requirements.