To work algebraically with exponential functions, we need to use the laws of exponents. You should

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1 Prealulus: Exponential and Logisti Funtions Conepts: Exponential Funtions, the base e, logisti funtions, properties. Laws of Exponents memorize these laws. To work algebraially with exponential funtions, we need to use the laws of exponents. You should If x and y are real numbers, and b > 0 is real, then 1. b x b y = b x+y 2. b x b y = bx y 3. (b x ) y = b xy Exponential Funtions A funtion of the form f(x) = a b x is an exponential funtion, where a 0 and b > 0, b 1 are real numbers. The number b is alled the base of the exponential funtion. Note: The differene between a monomial and an exponential is where the variable is. monomial: f(x) = x 3 has variable x in base. exponential: f(x) = 3 x has variable x in the exponent. onstant: f(x) = 3 π is a onstant funtion (no x). more ompliated funtion: f(x) = x x has an x in the base and in the exponent (you will see these in alulus). Exponential funtions are defined for all real numbers. For some real numbers, it is easy to figure out what the exponential funtion is. Consider f(x) = 3 x, whih is an exponential funtion. Evaluate at an integer: f(4) = 3 4 = = 81. Evaluate at zero: f(0) = 3 0 = 1. Evaluate at negative integer: f( 4) = 3 4 = = Evaluate at a rational number: f( 3/2) = 3 3/2 = 1 3 3/2 = However, we annot so easily figure out what an exponential funtion is when we evaluate it at a irrational number, for example what is f(π) = 3 π. We an determine the value of the number 3 π by a proedure whih will get us as lose as we want to the number: 3 π 3 3 = 27 3 π 3 31/10 = π 3 314/100 = π / = π In this way, we an evaluate an exponential at any real value for x. The sketh of the exponential funtion looks like: Page 1 of 6

2 Prealulus: Exponential and Logisti Funtions Skethes with different bases: Solid: y = 2 x, Dashed: y = 3 x, DotDashed: y = 4 x Solid: y = (1/2) x, Dashed: y = (1/3) x, DotDashed: y = (1/4) x Growth or Deay (Inreasing or Dereasing) If a > 0 and b > 1, then f(x) = a b x is inreasing and alled an exponential growth funtion (b is alled the growth fator). If a > 0 and 0 < b < 1, then f(x) = a b x is dereasing and alled an exponential deay funtion (b is alled the deay fator). This is also ( sometimes ) stated the following way, sine for b > 1 the quantity 1/b will be less than one, so we have exponential x 1 deay with = (b 1 ) x = b x, and we an say: b If a > 0 and b > 1, then a b x is inreasing, and a b x is dereasing. Example Determine the exponential funtion that passes through the points (0, 5) and (2, 16). The general exponential funtion is f(x) = a b x. We an use the two points we are given to determine the two onstants a and b. so a = 5. f(0) = 5 = a b 0 = a f(4) = 16 = 5 b 2 You might want to review the Properties of Exponents on page 8. equation. We have to use these properties to solve this 16 = 5 b 2 16 = b 2 5 ( ) 1/2 16 = (b 2 ) 1/ = b The exponential funtion that passes through the two points is y = f(x) = 5( 16/5) x. We do not use the negative root for b sine for exponential funtions the base must be greater than zero. Page 2 of 6

3 Prealulus: Exponential and Logisti Funtions The base e There is a preferred base in alulus (you will see why in alulus). It is an irrational number e = In alulus you will learn why this base is preferred to other bases. In alulus, you will also see where the definition of the number e e = x ( ) x x omes from, although it an be motivated using the idea of ontinuously ompounded interest. For now, remember that e is simply an irrational number, and it is the preferred base for the exponential funtion. The Natural Exponential Funtion f(x) = a e kx If a > 0 and k > 0, this is an exponential growth funtion. If a > 0 and k < 0, this is an exponential deay funtion. One we learn about logarithms, we shall see that we an always onvert from a general base b to base e using b x = e kx. Logarithms will allow to determine the value of k for a given b, but even without logarithms we an see this is true using the rules of exponents: f(x) = b x (rewrite b = e k for some value of k) = (e k ) x = e kx For this reason, I tend to fous on y = e kx for exponential growth and y = e kx for exponential deay, although we shall see that when modeling with exponentials it is often more benefiial to use a different base. The exponential funtion an be transformed using our typial transformation tehniques (for example, y = 2 x+3 is y = 2 x shifted to the left by three units). Page 3 of 6

4 Prealulus: Exponential and Logisti Funtions Example Sketh y = ae kx where a > 0 and k > 0 are real numbers. Basi Funtion y = f(x) = e x y = f( x) = e x reflet about y-axis y = f( kx) = e kx horizontal ompression by fator of k y = af( kx) = ae kx vertial streth by a fator of a From the sketh we an analyze the behaviour of the funtion f(x) = ae kx where a > 0 and k > 0: Range: y (0, ) Bounded below, but not above No loal extrema Horizontal Asymptote: y = 0 No Vertial Asymptotes Dereasing x ae kx = and x ae kx = 0 Logisti Funtions Both logisti and exponential funtions are used to model population growth. The problem with exponential growth is that it is unbounded, and at some point the model will no longer predit what is atually happening sine the amount of resoures available is finite and exponential growth annot ontinue on a purely physial level. If a, b,, and k are positive onstants, and b < 1, then a logisti growth funtion an be written as f(x) = 1 + a b x = 1 + a e kx Logisti growth is bounded above and below, so it is a more realisti model of population growth. Using the sketh of y = ae kx from the previous example, we an atually onstrut the sketh of the logisti equation. Page 4 of 6

5 Prealulus: Exponential and Logisti Funtions First, let s determine the end behaviour of the logisti funtion: x x 1 + a e kx = 1 + a 0 = 1 + a e kx = 1 + a = 0 So the logisti funtion has two horizontal asymptotes, y = 0 and y =. Does the logisti funtion have any vertial asymptotes? The answer would be yes if 1 + a e kx = 0 sine then we would have division by zero. This would happen if ae kx = 1, whih never happens sine the range of ae kx is (0, ). So the logisti funtion has no vertial asymptotes, and it is ontinuous on the domain x (, ). The logisti funtion has no zeros, sine 0. The y-interept of the logisti funtion is f(0) = 1 + a. Putting all this information together, we an get a sketh of the logisti funtion In partiular, here is a sketh of f(x) = e 6x : If you ontinue with math, you will see improvements to the logisti growth that take into aount extintion of a speies (this is typially done in a differential equations ourse). Page 5 of 6

6 Prealulus: Exponential and Logisti Funtions Properties of y = e x Range: y (0, ) Bounded below, but not above No loal extrema Horizontal Asymptote: y = 0 No Vertial Asymptotes Inreasing x ex = 0 and x ex = Properties of y = e x Range: y (0, ) Bounded below, but not above No loal extrema Horizontal Asymptote: y = 0 No Vertial Asymptotes Dereasing x e x = and x e x = 0 Properties of y = 1 ae bx Range: y (0, ) Bounded above and below No loal extrema Horizontal Asymptotes: y = 0 and y = No Vertial Asymptotes Inreasing x = 0 and 1 ae bx x 1 ae bx = Page 6 of 6