Math 102 Answers

1. x x Sketch the graph of f(x) = e ! 2. Label the intercept and asymptote. Sketch the graph of g(x) = ln (x+2) . Label the intercept and asymptote. x f(x) = e !2 x The graph of f(x) = e ! 2 is the graph of y = e shifted down 2 units. x y = e has a horizontal asymptote at the x-axis, y = 0. x f(x) = e ! 2 has a horizontal asymptote at y = !2. The graph of g(x) = ln (x+2) is the graph of y = ln (x) shifted left 2 units. y = ln (x) has a vertical asymptote at the y-axis, x=0, so g(x) = ln (x+2) has a vertical asymptote at x = !2. 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 (3x+2) + ln (x) = 0 ln ( x (3x+2) ) = y=x x=!2 To solve a logarithmic equation, we use exponential function. But first it would help to get one ln expression. 0 Now we are ready to take the exponential function on each side eln (x (3x+2) ) = e0 (x (3x+2) ) = 1 3x2 + 2x ! 1 = 0 (3x!1) ( x+1 ) = 0 x = a or !1 x = a 3. 5 x— 3 = 2 x . Solve for x: ln 5 x— 3 = ln 2 x (x ! 3) ln 5 and simplify.... Resulting in a quadratic equation which we may solve by factoring, or by quadratic formula And now, you must check these “solutions”. Since ln (!1) does not exist, !1 does not solve the equation. a does: ln (3) + ln (a) = ln (3) ! ln (3) = 0 To solve an exponential equation, we use a logarithm. Taking ln of both sides... Using the logarithmic property ln x p = p ln x ... = x ln 2 x ln 5 ! 3 ln 5 = x ln 2 Keep in mind that you are solving for x x ln 5 ! x ln 2 = 3 ln 5 x (ln 5 ! ln 2) = 3 ln 5 x Solve for x: = 8 x— 6 One division step away from the answer... 3 ln 5 ln 5 ! ln2 )))))))))) = 2x (2 3) x— 6 = 2 x 2 3 x— 18 = 2 x 3x ! 18 = x x = 9 can be solved in the same manner as above, or we can take advantage of the fact that 2 3 = 8 to get And since exponential functions are 1-1, we know this follows: 1 4. Find inverse function: f(x) = ))))) . 3x !1 x+1 3x ! 1 ))))) f(x) = y = First we interchange x & y then we solve for y.... We can use the method shown at right, or... y+1 )))))) 3y ! 1 x = observe that f multiplies by 3, subtracts 1, then takes reciprocal multiply both sides by (3y !1) 3xy ! x = y + 1 so f )1 must take reciprocal, add 1, then divide by 3. That is, f)1 (x) 1 )) + 1 = x ))))) = then add x, add !y 3xy ! y = 1+x 3x )))))) y = 1+x 1+x 3x ! 1 )))))) f)1 (x) = 1+x 3x ! 1 )))))) = f(x) 3 (10) 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 essential to notice we can combine them .... 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 400 = 4 e r2 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. 7. According to the rational zeroes theorem, only certain rational numbers may be zeroes of the polynomial 3 2 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 3 2 function P(x) = 2x — 5x + 5x + 4. ½ 2 2 !5 1 !4 5 !2 3 !½ 4 3/2 5½ !5 !1 !6 2 2 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 ) ² This helps finish #9. = (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 ± %& !& 7 Using the quadratic formula, Q(x) = 0 when x = ))))))) 2 3 = 3 ± %& 7i 2 ))))))) 2 The zeroes of P(x) = 2x — 5x + 5x + 4 are !1'2 and 3'2 ! r%%7 i'2 and 3'2 + r%%7 i'2 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.