Exponential models #1 - Questions and Answers

Question 1 Which set of ordered pairs could be generated by an exponential function? A. (1,1),(2,½),(3,¼),(4,⅛) B. (1,1),(2,½),(3,⅙),(4,⅛) C. (1,½),(2,¼),(3,⅛),(4,1/16) D. (1,½),(2,1),(3,¾),(4,7/16) Answer An exponential function has the property that the y-values are multiplied by some common factor a when x increases by 1. We note that each set of ordered pairs contains 1,2,3 and 4 as x-values, which are increments of 1. Let us next determine the ratio of each pair of consecutive y-values for each option. Answer option A ½/1 = 1/2 = 0.5 ¼/½ = ½ = 0.5 ⅛/¼ = ½ = 0.5 Answer option B ½/1 = 0.25 ⅙/½ = ⅓ ≈ 0.4444 ⅛/⅙ = ¾ ≈ 0.5625 Answer option C ¼/½ = ½ = 0.5 ⅛/¼ = ½ = 0.5 1/16 / ⅛ = ½ = 0.5 Answer option D 1/16 / ½ = 2/3 ≈ 0.6667 ¾ / 1 = 3/4 = 0.75 7/16 / ¾ = 7/12 ≈ 0.5833 We note that the three ratios are only constant for answer option C and thus answer option C could be generated by an expo. function. ANSWER: C. (1,½),(2,¼),(3,⅛),(4,1/16) Question 2 Find The Exponential Function f(x) = aˣ Whose Graph Is Given. Find the EXPONENTIAL FUNCTION F(x) = Aˣ WHOSE GRAPH is given F(x) = ? Answer 1 In this question, we have given point on curve (-2,9) and f(x) = aˣ So, plug in values 9 = a⁻² 9 is 3² and (⅓)⁻² = 9 So, correct graph is y = (⅓)ˣ Answer 2 Calculation: In this question, we have given point on curve (-2,9) and f(x) = aˣ So, plug in values 9 = a⁻² 9 is 3² and (⅓)⁻² = 9 So, correct graph is y = (⅓)ˣ Question 3 Solve eˣ = 0 Answer 1 That is 0 is the only value that eˣ cannot take Thus, we can say, that there is no x such that eˣ = 0 Answer 2 There is no x for eˣ = 0 Question 4 Find the derivative of y = e⁵ˣ Answer 1 The derivative of eˣ is simply eˣ. However, x has a coefficient, so you have to use the chain rule. If y = e⁵ˣ by the chain rule, the derivative will be equal to the derivative of e⁵ˣ with respect to 5x, multiplied by the derivative of 5x with respect to x. dy/dx = d/d(5x) [e⁵ˣ] x d/dx [5x] dy/dx = e⁵ˣ x 5 dy/dx = 5e⁵ˣ Answer 2 Apply the cahin rule: d/dx (e⁵ˣ) = e⁵ˣ d/dx (5x) d/dx (5x) = 5 e⁵ˣ x 5 = 5e⁵ˣ Question 5 The fox population in a certain region has a continuous growth rate of 6 percent per year. It is estimated that the population in the year 2000 was 18900. (a) Find a function that models the population t years after 2000 (t=0 for 2000). Hint: Use an exponential function with base e. (b) Use the function from part (a) to estimate the fox population in the year 2008. Answer 1 The fox population in a certain region has a continuous growth rate of 6 percent per year. It is estimated that the population in the year 2000 was 18900. Calculation: The general form of an exponential equation is, P(t) = P₀e^rt Here, P₀ = 18900 r = 6% = 6/100 = 0.06 a) Therefore, a function that models the population t years after 2000 is P(t) = 18900e^0.06t b) For 2000, t=0 Therefore for 2008, t=8 Therefore, the estimation of fox population in the year 2008 is P(8) = 18900e^0.06(8) P(8) ≈ 30544 Answer 2 (a) We know that the population has a continuous growth rate of 6 percent per year, which can be expressed as a growth factor of 1 + 0.06 = 1.06 per year. Since we're starting from a population of 18,900 in the year 2000, we have the initial condition P(0) = 18,900. Using these pieces of information, we can write the exponential function as: P(t) = P(0) * (1 + r)ᵗ where r is the growth rate per year. Substituting the given values, we have: P(t) = 18900 * (1 + 0.06)ᵗ Simplifying further: P(t) = 18900 * (1.06)ᵗ Therefore, the function that models the population t years after 2000 is P(t) = 18900 * (1.06)ᵗ. (b) To estimate the fox population in the year 2008 (8 years after 2000), we can substitute t = 8 into the function we found in part (a): P(8) = 18900 * (1.06)⁸ Calculating the value: P(8) ≈ 18900 * (1.06)⁸ ≈ 18900 * 1.593848 ≈ 30,144.43 Therefore, the estimated fox population in the year 2008 is approximately 30,144 (rounded to the nearest whole number). Answer 3 (a) To model the population growth, we can use the formula for exponential growth: P(t) = P₀ * e^rt where: P(t) represents the population at time t, P₀ is the initial population (at t=0), e is the base of the natural logarithm (approximately 2.71828), r is the growth rate as a decimal (6 percent = 0.06), and t is the time in years. Given that the population in the year 2000 was 18900 (at t=0), we can substitute the values into the equation: P(t) = 18900 * e^0.06t (b) To estimate the fox population in the year 2008 (t=8), we can substitute t=8 into the function from part (a): P(8) = 18900 * e^0.068 Now we can evaluate the expression to find the estimated population in the year 2008. Answer 4 Result: (a) P(t) = 18900 * e^0.06t (b) 30,582 (a) To model the fox population, we can use an exponential function with base e. Let's denote the population at time t years after 2000 as P(t). We know that the continuous growth rate is 6 percent per year, which can be written as a decimal as r = 0.06. Since the base of our exponential function is e, the general form is P(t) = P₀e^rt, where P₀ is the initial population at t = 0. Given that the population in the year 2000 was 18,900, we have P₀ = 18900. Substituting these values into the equation, we get: P(t) = 18900 * e^0.06t (b) To estimate the fox population in the year 2008, we need to find the value of P(8) since 2008 is 8 years after 2000. Substituting t = 8 into the equation derived in part (a), we have: P(8) = 18900 * e^0.068 To evaluate this expression, we can use the approximate value of e as 2.71828. Calculating this, we find: P(8) = 18900 * e^0.48 ≈ 18900 * 1.618742 ≈ 30582 Therefore, the estimated fox population in the year 2008 is approximately 30,582.