Chap 6 Special Functions

Chap 6 Special Functions - FIRST YEAR CALCULUS W W L CHEN c...

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Unformatted text preview: FIRST YEAR CALCULUS W W L CHEN c W W L Chen, 1982, 2005. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 6 SOME SPECIAL FUNCTIONS 6.1. Exponential Functions In this section, we construct a class of functions of the form fa : R → R, where for every x ∈ R, fa (x) = ax . Here a > 0 denotes a positive real constant. Let us state very carefully what we mean by ax . We would like to define ax appropriately so that ax+y = ax ay for every x, y ∈ R. To do so, we must have ax+0 = ax a0 . This forces us to write (1) Also, it seems reasonable to write (2) an = a . . . a n times a0 = 1. for every n ∈ N. Next, it is clear that it is necessary to define, for every p, q ∈ N, (3) and (4) Chapter 6 : Some Special Functions y = a1/q > 0 if and only if y q = a, ap/q = (a1/q )p . page 1 of 4 First Year Calculus c W W L Chen, 1982, 2005 Note that (2)–(4) give ax for every x ∈ Q+ , the set of all positive rational numbers. Our definition is now extended to cover the set all negative rational numbers Q− by (5) ax = 1 a−x for every x ∈ Q− . Hence we have, by (1)–(5), defined ax for every x ∈ Q. The question that remains is how we define ax when x is irrational. Without giving all the details, we claim that it is possible to define ax for all irrational numbers x so that the function fa (x) = ax is continuous and differentiable everywhere in R. Now let us consider the derivative fa (x). Clearly fa (x) = lim Let us write c(a) = lim ah − 1 . h→0 h ay − ax ax+h − ax ah − 1 = lim = fa (x) lim . y →x y − x h→0 h→0 h h Numerical evidence suggests that c(2) < 1 and c(3) > 1. Indeed, we have c(2) < 0.7 and c(3) > 1.09. In fact, it can be shown that the function c(a) is continuous at every a ∈ [2, 3], so it follows from the Intermediate value theorem that there exists e ∈ (2, 3) such that c(e) = 1. With this number e, we have the function f : R → R, where for every x ∈ R, f (x) = ex . The results below are easy consequences of our discussion. PROPOSITION 6A. The function f : R → R, defined for every x ∈ R by f (x) = ex , has the following properties: (a) f (x) > 0 for every x ∈ R, and f (0) = 1. (b) f (x1 + x2 ) = f (x1 )f (x2 ) for every x1 , x2 ∈ R. (c) f (x) is differentiable, and f (x) = f (x) for every x ∈ R. (d) f (x) is strictly increasing in R; in other words, f (x1 ) < f (x2 ) whenever x1 < x2 . (e) f (x) → 0 as x → −∞. (f ) f (x) → +∞ as x → +∞. 6.2. The Exponential and Logarithmic Functions It is easy to see that the function considered in Proposition 6A is one-to-one, in view of part (d). On the other hand, the function is not onto, in view of part (a). However, this “mishap” can be corrected easily by changing the codomain to R+ , the set of all positive real numbers. So let us change the codomain. PROPOSITION 6B. The function exp : R → R+ , defined for every x ∈ R by exp(x) = ex , is one-to-one and onto. Definition. The function exp(x) is usually called the exponential function. It now follows from Proposition 2A that the exponential function exp : R → R+ has an inverse function. This is known as the logarithmic function, and denoted by log : R+ → R. Hence y = exp(x) Chapter 6 : Some Special Functions if and only if x = log(y ). page 2 of 4 First Year Calculus c W W L Chen, 1982, 2005 The results below are easy consequences of our discussion. PROPOSITION 6C. The logarithmic function log : R+ → R has the following properties: (a) log(y ) > 0 for every y > 1, log(y ) < 0 for every positive y < 1, and log(1) = 0. (b) log(y1 y2 ) = log(y1 ) + log(y2 ) for every y1 , y2 ∈ R+ . (c) log(y ) is differentiable, and log (y ) = 1/y for every y ∈ R+ . (d) log(y ) is strictly increasing in R+ ; in other words, log(y1 ) < log(y2 ) whenever 0 < y1 < y2 . (e) log(y ) → −∞ as y → 0+. (f ) log(y ) → +∞ as y → +∞. The only difficult part is (c). Here we can use the result dx/dy = 1/(dy/dx). Then if x = log(y ), then y = exp(x); hence dy = exp(x) = y dx 2 and dx 1 =. dy y Example 6.2.1. Consider the function f (x) = ex −2x . The graph does not intersect the x-axis, and intersects the y -axis at the point (0, 1). Also, f (x) → +∞ as x → +∞ and as x → −∞. On the other hand, it follows from the Chain rule that f (x) = (2x − 2)ex 2 −2x < 0 if x < 1, = 0 if x = 1, > 0 if x > 1. Hence there is a stationary point at x = 1. Also the function is decreasing when x < 1 and increasing 2 when x > 1. Now f (x) = ((2x − 2)2 + 2)ex −2x > 0 always. It follows that the function has a local minimum at x = 1. Furthermore, the slope of the tangent is always increasing. Example 6.2.2. Consider the (even) function f (x) = log(x2 + 1). Chapter 6 : Some Special Functions page 3 of 4 First Year Calculus c W W L Chen, 1982, 2005 The graph intersects the coordinate axes at the point (0, 0). Also, f (x) → +∞ as x → +∞ and as x → −∞. On the other hand, it follows from the Chain rule that f (x) = 2x x2 + 1 < 0 if x < 0, = 0 if x = 0, > 0 if x > 0. Hence there is a stationary point at x = 0. Also the function is decreasing when x < 0 and increasing when x > 0. Now < 0 if x < −1, 2 − 2x2 = 0 if x = −1, f (x) = 2 > 0 if −1 < x < 1, (x + 1)2 = 0 if x = 1, < 0 if x > 1. It follows that the function has a local minimum at x = 0. Also it has points of inflection at x = −1 and at x = 1. Furthermore, the slope of the curve is decreasing in the intervals (−∞, −1) and (1, ∞), and increasing in the interval (−1, 1). Problems for Chapter 6 1. Show that the function f (x) = is not continuous at x = 0. 2. Show that the function f (x) = is continuous everywhere. 3. Find a largest domain, the corresponding range and derivative of each of the following functions: a) f (x) = 4 sin−1 (5x) b) f (x) = 1 cos−1 (x2 ) c) f (x) = tan−1 (x2 − 1) 2 4. Let f (x) = x+1 . Find the inverse function of f (x) if it exists. x−1 e−1/|x| 0 if x = 0, if x = 0, e−1/x 0 if x = 0, if x = 0, 5. Does the function y = x3 − 2 have an inverse function on the real line? Give your reasons. If yes, then find also the inverse function. 6. For each of the following functions, find the inverse if it exists: x+2 a) f (x) = x3 + 1 b) f (x) = 2x − 1 c) f (x) = x2 − 4x + 3 on domain {x ∈ R : x ≥ 2} Chapter 6 : Some Special Functions page 4 of 4 ...
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This note was uploaded on 02/01/2009 for the course MATH 342412 taught by Professor Staff during the Spring '09 term at UCSD.

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