Calculus for the Life Sciences I

Question 1. [5 points] Let f (x) = 1 − x 2 and g (x) = ln x. Find (a)[2 points] f (g (x)) and its domain; Solution: ¡ ¢ f g (x) = 1 − (ln x)2 , its domain is (0, ∞). (b)[3 points] g ( f (x)) and its domain. Solution: ¡ ¢ g f (x) = ln(1 − x 2 ), from which, we have 1 − x 2 > 0 ⇒ x 2 < 1 ⇒ |x| < 1 ⇒ −1 < x < 1. Question 2. [10 points] 500 g of iodine 131 is decaying exponentially. After 3 days 386 g of iodine 131 is left. (a)[6 points] Write an exponential equation to express the amount of iodine 131 as a function of time; Solution: y = y 0 e kt , when t = 0, y = 500, so y 0 = 500, i.e., y = 500e kt ; when t = 3, y = 386. 386 = 500e 3k ⇒ 386 ln 0.772 = e 3k ⇒ e 3k = 0.772 ⇒ 3k = ln 0.772 ⇒ k = 500 3 k ≈ −0.0863 Hence, y = 500e −0.0863t (b)[4 points] Use your answer from part (a) to find the half-life of iodine 131. Solution: ln 21 1 1 1 −0.0863t −0.0863t y0 = y0e ⇒ =e ⇒ ln = −0.0863t ⇒ t = ≈ 8.0 2 2 2 −0.0863 The half-life is about 8.0 days. Question 3. [15 points] Use properties of Logarithms: p 2 x +1 (a)[5 points] Expand log (x+2) (x+1)2 (x−3) Solution: p p ¢ ¡ ¢ ¡ (x + 2) x 2 + 1 2 + 1 − log (x + 1)2 (x − 3) x log = log (x + 2) (x + 1)2 (x − 3) p ¡ ¢ = log(x + 2) + log x 2 + 1 − log(x + 1)2 + log(x − 3) p = log(x + 2) + log x 2 + 1 − log(x + 1)2 − log(x − 3) 1 = log(x + 2) + log(x 2 + 1) − 2 log(x + 1) − log(x − 3). 2 (b)[5 points] Find the integer value of (log3 81)(log4 64) Solution: (log3 81)(log4 64) = log3 34 log4 43 = (4 log3 3)(3 log4 4) = 4 · 3 = 12. (c)[5 points] Solve for x: log2 x = 2 − log2 (4 − x) Solution: log2 x + log2 (4 − x) = 2 so that log2 x(4 − x) = 2 and 4x − x 2 = x(4 − x) = 22 = 4. So (x − 2)2 = x 2 − 4x + 4 = 0 ⇒ x = 2. Question 4. [10 points] Let f (x) = 2 cos(πx) − 1. (a)[4 points] Find f (0), f ( 13 ), f ( 12 ), f (1); Solution: f (0) = 2 cos 0 − 1 = 2 − 1 = 1 π 1 1 f ( ) = 2 cos( ) − 1 = 2 − 1 = 1 − 1 = 0 3 3 2 1 π f ( ) = 2 cos( ) − 1 = 0 − 1 = −1 2 2 f (1) = 2 cos(π) − 1 = 2(−1) − 1 = −3. (b)[4 points] Graph f (x) over an interval of one period; Solution: (c)[2 points] Find the maximum and minimum values of the function. Solution: max = 2 and min = −2. 5 Question 5. [15 points] Find the limits x 2 − 3x − 10 x→5 x −5 (a) [3 points] lim Solution: x 2 = 3x − 10 (x − 5)(x + 2) = lim = lim (x + 2) = 5 + 2 = 7. x→5 x→5 x→5 x −5 x −5 lim x2 − 9 x→−3 x 2 + x − 6 (b) [3 points] lim Solution: x2 − 9 (x − 3)(x + 3) (x − 3) −3 − 3 6 = lim = lim = = . x→−3 x 2 + x − 6 x→−3 (x − 2)(x + 3) x→−3 (x − 2) −3 − 2 5 p x −6 (c) [3 points] lim x→36 x − 36 lim Solution: p p x −6 x −6 1 1 1 1 = lim p = . lim = lim p =p = p x→36 x − 36 x→36 ( x − 6)( x + 6) x→36 x + 6 36 + 6 6 + 6 12 x 2 + 2x − 5 x→∞ 3x 2 + 2 (d) [2 points] lim Solution: x 2 + 2x − 5 lim = lim x→∞ 3x 2 + 2 x→∞ x2 x2 + 2x − x52 x2 3x 2 x2 + x22 = lim 1 + x2 − x52 x→∞ 3+ 2 x2 1+0−0 1 = 3+0 3 = 2x 2 − 1 x→∞ 3x 4 + 2 (e) [2 points] lim Solution: 2x 2 − 1 lim = lim x→∞ 3x 4 + 2 x→∞ 2x 2 x4 3x 4 x4 − x14 + x24 = lim x→∞ 2 x2 − x14 3+ 2 x4 = 0−0 0 = = 0. 3+0 3 x 4 − x 3 − 3x x→∞ 7x 2 + 9 (f) [2 points] lim Solution: x 4 − x 3 − 3x lim = lim x→∞ x→∞ 7x 2 + 9 x4 x4 3 − xx 4 − 3x x4 7x 2 x4 + x94 = lim x→∞ 1 − x1 − x33 7 x2 + 9 x4 1−0−0 1 = . 0+0 0 = Division by 0 is undefined, so this limit does not e xist. By examining what happens as larger 4 and larger values of x are put into the function, we see that limx→∞ x −x 3 −3x = 7x 2 +9 ∞. Question 6. [6 points] Let  if x < 0  0 2 x − 5x if 0≤x ≤5 f (x) =  5 if x > 5 (a)[3 points] Graph the above function; Solution: (b)[1 point] Find all values of x where the function is discontinuous; Solution: f (x) is discontinuous at x = 5. (c)[2 points] Find the limit from the left and from the right at any values of x found in part (b). Solution: limx→5− f (x) = limx→5− (x 2 − 5x) = 52 − 5 · 5 = 0. limx→5+ f (x) = limx→5+ 5 = 5. Question 7. [10 points] Epidemiologists estimate that t days after the flu begins to spread in some college town, the percent of the population infected by the flu is approximated by p(t ) = t 2 + t for 0 <= t <= 5. (a)[4 points] Find the average rate of change of p with respect to t over the interval from 1 to 4 days; Solution: The average rate is equal to p(4) − p(1) (42 + 4) − (12 + 1) 20 − 2 18 = = = = 6. 4−1 4−1 3 3 (b)[6 points] Find the instantaneous rate of change of p with respect to t at t = 3. Solution: Consider limh→0 We have p(t +h)−p(t ) h at t = 3. p(3 + h) − p(3) h→0 h (3 + h)2 + (3 + h) − (32 + 3) = lim h→0 h 2 2 3 + 6h + h + 3 + h − 12 = lim h→0 h h 2 + 7h = lim = lim (h + 7) h→0 h→0 h = 7. lim 2 Question 8. [12 points] Let f (x) = 1 − x2 . (a)[6 points] Use the definition of the derivative to find f 0 (x); Solution: (1 − f (x + h) − f (x) f (x) = lim = lim h→0 h→0 h 0 2 − (x+h) 2 x2 2 2 (x+h)2 ) − (1 − x2 ) 2 h 2 = lim 2 1 − (x+h) − 1 + x2 2 h→0 h x 2 − (x + h)2 x 2 − x 2 − 2xh − h 2 = lim h→0 h→0 h→0 h 2h 2h 2 −2xh − h −2x − h = lim = lim = −x. h→0 h→0 2h 2 = lim + = lim (b)[4 points] Find the equation of the tangent line to the graph at x = −2; Solution: 2 0 y − y 0 = m(x − x 0 ). x 0 = −2 so that y 0 = f (x 0 ) = 1− (−2) 2 = 1−2 = −1. m = f (x 0 )− x 0 = 2. Also, y − (−1) = 2(x − (−2)) ⇒ y + 1 = 2(x + 2) ⇒ y = 2x + 3. (c)[2 points] Sketch the tangent line from part (b) on the graph below. Find the y-intercept. Solution: The y-intercept is 3. Question 9. [BONUS: 9 points] Sketch the graph of the derivative for the function shown. y -6 -4 2 -2 -1 -2 -3 You can sketch your answer in the following graph: Solution: 4 6 x