1. State whether the function is an exponential growth or exponential decay, and describe its end behaviour using limits.

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1 Questions 1. State weter te function is an exponential growt or exponential decay, and describe its end beaviour using its. (a) f(x) = 3 2x (b) f(x) = 0.5 x (c) f(x) = e (d) f(x) = ( ) x Matc te function wit its grap (a) y = e x (b) y = e x (c) y = e x (d) y = e x Grap te function f(x) = 3e x + 2 by transforming te basic function y = e x, and analyze it for domain, range, continuity, increasing or decreasing beaviour, symmetry, boundedness, extrema, asymptotes, and end beaviour. 4. Sketc te Logistic function f(x) =. Label te y-intercept, and analyze te function for domain, range, 1 + 3e 2x continuity, increasing or decreasing beaviour, symmetry, boundedness, extrema, asymptotes, and end beaviour. 5. (calculus preview) In tis problem we will investigate te instantaneous rate of cange of te exponential function f(x) = e 2x 6. Simplify te average rate of cange = f(x + ) f(x) so tat it involves a function of x times te quantity e2 1. f(x + ) f(x) e 2 1 Te instantaneous rate of cange =. Explain wy can you not evaluate substitution. Evaluate 0 e 2 1 using te following ideas. by direct Page 1 of 7

2 First, sketc e 2 using a calculator near = 0. Pick a domain tat makes te function e 2 look like a straigt line, and ten determine te equation of te straigt line a + b for wic e 2 a + b. Ten, evaluate te it using e 2 1 (a + b) 1 Finally, write down an expression for te instantaneous rate of cange of e 2x 6. Page 2 of 7

3 Solutions 1. State weter te function is an exponential growt or exponential decay, and describe its end beaviour using its. (a) f(x) = 3 2x = ( ) 2x 1 = 3 9 Base b = 1/9 < 1, so we ave exponential decay. (b) f(x) = 0.5 x = 2 f(x) = 0 and f(x) =. Base b = 1/2 < 1, so we ave exponential decay. (c) f(x) = e f(x) = 0 and f(x) =. Base b = 1/e < 1, so we ave exponential decay. f(x) = 0 and f(x) =. Alternately, rewrite as f(x) = = (e 1 ) x = e x, so we ave exponential decay. e (d) f(x) = ( ) x 1 = 4 x 4 Base b = 4 > 1, so we ave exponential growt. f(x) = and f(x) = Matc te function wit its grap (a) y = e x (b) y = e x (c) y = e x (d) y = e x + 2 (d) (a) (b) (c) Page 3 of 7

4 3. Grap te function f(x) = 3e x + 2 by transforming te basic function y = e x, and analyze it for domain, range, continuity, increasing or decreasing beaviour, symmetry, boundedness, extrema, asymptotes, and end beaviour. Here are te algebraic representations of te transformations: Basic function: y = g(x) = e x. Flip about te y-axis: y = g( x) = e x. Flip about te x-axis: y = g( x) = e x. Stretc vertically by a factor of 3: y = 3g( x) = 3e x. Sift up 2 units: y = 3g( x) = 3e x + 2 = f(x). Here are te sketces: Page 4 of 7

5 Domain: x R Range: y (, 2) (it never reaces y = 2). Continuity: continuous for all x Increasing-decreasing beaviour: increasing for all x Symmetry: none Boundedness: bounded above Local Extrema: none Horizontal Asymptotes: y = 2 Vertical Asymptotes: none End beaviour: x ( 3e x + 2) = and x ( 3e x + 2) = 2 4. Sketc te Logistic function f(x) =. Label te y-intercept, and analyze te function for domain, range, 1 + 3e 2x continuity, increasing or decreasing beaviour, symmetry, boundedness, extrema, asymptotes, and end beaviour. We know te basic sape of a logistic function, so to get te sketc we just ave to figure out te value of te upper vertical asymptote. We can use te fact tat x e 2x = 0 to do tis. x Te y-intercept is at 1 + 3e 2x = 1 + 3(0) = f(0) = 1 + 3e 0 = 1 + 3(1) = 6. Domain: x R Range: y (0, ) Continuity: continuous for all x Increasing-decreasing beaviour: increasing for all x Symmetry: none Boundedness: bounded above and below Local Extrema: none Horizontal Asymptotes: y = 0 and y = Vertical Asymptotes: none End beaviour: x = 0 and 1 + 3e 2x x = 1 + 3e 2x Page 5 of 7

6 5. average rate of cange = f(x + ) f(x) = e2(x+) 6 e 2x 6 = e2x+2 6 e 2x 6 = e2x 6 e 2 e 2x 6 = e 2x 6 e2 1 instantaneous rate of cange = f(x + ) f(x) = e 2x 6 e2 1 = e 2x 6 e 2 1 If we try to use direct substitution to evaluate tis it, we get an indeterminant form: e 2 1 = e0 1 = 1 1 = Remember, an indeterminant form just means we don t know wat te number is, and ave to do some more work to figure it out. You will learn ow to treat problems like tis in calculus witout making an approximation like we do below. For now, we ave to do some estimating to get an answer. Here is te sketc: It looks like te e 2 passes troug two points, (0, 1) and (0.05, 1.1). Te first point it passes troug exactly, te second it only comes close too. From ere on we are using an approximation. Page 6 of 7

7 Te equation of te straigt line troug two points is: y y 1 = 1 y 2 y y 1 0 = y 1 = y = = So we ave sown e wen 0. Let s use tis to work out te instantaneous rate of cange: instantaneous rate of cange = e 2x 6 e 2 1 e 2x 6 (2 + 1) 1 = e 2x 6 2 = e 2x = e 2x 6 (2) = 2e 2x 6 Terefore, we ave sown tat te instantaneous rate of cange of e 2x 6 is 2e 2x 6. You will learn a way of sowing tis using derivatives wen you take calculus, tat does not rely on te approximations we ave made. Page 7 of 7