Exponential models #1 - Q&A

Question 1 Which set of ordered pairs could be generated by an exponential function? Α. (1,1)(2)(3,4) (4) 8. (1,1)(2,1)(3,)(4) C. (1,)(2)(3)(4) D. (1,)(2)(3)(4) Answer 1 Question 2 Find the EXPONENTIAL FUNCTION FIX)=A" WHOSE GRAPH is given F(x)=? y (29) Answer 1 in this question, we have given point on were (-2,9) 2kappa * df(x) = e ^ x So, ply in values g = Q ^ - 2 I are (9 So, corpeat grapes Answer 2 Calculation: In this question, we have given point on curve (-2,9) and f(x) = a ^ 2 So, plug in values 9 = a ^ - 2 9is * 3 ^ 2 * and * (1/3) ^ - 7 = 9 y = (1/3) ^ x Question 3 Solve e ^ x = 0 Answer 1 That is O is the only value that cannot take Thus, we can say, that there is no x such that c ^ 2 = 0 Answer 2 There is no x for e ^ x = 0 Question 4 Find the derivative of y = e ^ (5x) Answer 1 The derivative of is simply e. However, x has a coefficient, so you have to use the chain rule. If y = e ^ (5y) by the chain rule, the derivative will be equal to the derivative of it with respect to 5, multiplied by the derivative of 5 with respect to x d/dx (y) = d/dx ([5x] * e ^ (5x)) (ay)/(dx) = 5e ^ (5x) Answer 2 d/dx (e ^ (3x)) the cahin rule: = e ^ (5x) * d/dx (5x) d/dx (5x) = 5 x5 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)= Pe Here, P18900 r = 6% = 100 =0.06 a) Therefore, a function that models the population t years after 2000 is P(t) = 18900e ^ (0ts) b 100, t = 0 Therefore for 2008, t = 8 Therefore, the estimation of fox population in the year 2008 is P(8) 18900000(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 10.06 106 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 pleces of information, we can write the P(t)=P(0) (1+r) where r is the growth rate per year. Substituting the given values, we have exponential function as: P(t) = 1.09 - (1 + 0.06) ^ 2 Simplifying further: P(t) = 18900 - (1.06) ^ 1 Therefore, the function that models the population t years after 2000 is P(t) 18900-(1.06). (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(B) 18900-(1.06) 10900-1.593848 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)= Poxe where: P(t) represents the population at time t e is the base of the natural logarithm (approximately 2.71828), ris the growth rate as a decimal (6 percent = 0.06), and is the time in years Po is the initial population (att = 0) Given that the population in the year 2000 was 18900 (ott = 0) we can substitute the values into the equation: P(t) = 18900e ^ (0.08t) (the fox population in the year 2008 (t = 8) , we can substitute t = 81 into the function from part (a): P(0) = 18900e ^ (5.06g) Now we can evaluate the expression to find the estimated population in the year 2008. Answer 4 Result: (a)P(1) = 18900 (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 years after 2000 as P(r). We know that the continuous growth rate is 6 percent per year, which can be written as a decimal as = 0.06 Since the base of our exponential function is e, the general form is P(t) Pove", where Pe is the initial population at := 0. Given that the population in the year 2000 was 18,900, we have Po 18900. Substituting these values into the equation, we get: P(t)=18900 (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. Substitutingt t = 8 into the equation derived in part (a), we have: P(8) = 18 deg To evaluate this expression, we can use the approximate value of e as 2.71828. Calculating this, we find P(8) = 1 18900416900-1.618742 Z = 30582 Therefore, the estimated fox population in the year 2008 is approximately 30,582.