SalasSv_07_03 - 388 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS c In Exercises 2628 a function g is given(a Show that there is a number r in the indicated

SalasSv_07_03 - 388 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS...

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7.3 THE LOGARITHM FUNCTION, PART II Differentiation and Graphing We know that for x > 0, d dx (ln x ) = 1 x . Now suppose that u is a positive, differentiable function of x . Then (7.3.1) d dx (ln u ) = 1 u du dx . PROOF By the chain rule, d dx (ln u ) = d du (ln u ) du dx = 1 u du dx . For example, d dx [ln (1 + x 2 )] = 1 1 + x 2 · 2 x = 2 x 1 + x 2 for all x , and d dx [ln(1 + 3 x )] = 1 1 + 3 x · 3 = 3 1 + 3 x for all x > 1 3 . Example 1 Find the domain of f and find f ( x ) if f ( x ) = ln ( x 4 + x 2 ). SOLUTION For x to be in the domain of f , we must have x 4 + x 2 > 0, and thus we must have x > 0. The domain of f is the set of positive numbers. Before differentiating f , we make use of the special properties of the logarithm: f ( x ) = ln x 4 + x 2 = ln x + ln [(4 + x 2 ) 1 / 2 ] = ln x + 1 2 ln (4 + x 2 ). From this we have f ( x ) = 1 x + 1 2 · 1 4 + x 2 · 2 x = 1 x + x 4 + x 2 = 4 + 2 x 2 x (4 + x 2 ) .
7.3 THE LOGARITHM FUNCTION, PART II 389 Example 2 Sketch the graph of f ( x ) = ln | x | . SOLUTION The function, defined at all x = 0, is an even function: f ( x ) = f ( x ) for all x = 0. The graph has two branches: y = ln ( x ), x < 0 and y = ln x , x > 0. Each branch is the mirror image of the other. (Figure 7.3.1) Example 3 ( Important ) Show that y = In ( x ) y = In x x y y = In x Figure 7.3.1 (7.3.2) d dx (ln | x | ) = 1 x for all x = 0. SOLUTION For x > 0, d dx (ln | x | ) = d dx (ln x ) = 1 x . For x < 0, we have | x | = − x > 0, so that d dx (ln | x | ) = d dx [ln ( x )] = 1 x d dx ( x ) = 1 x ( 1) = 1 x . It follows that if u is a differentiable function of x , then where u ( x ) = 0, (7.3.3) d dx (ln | u | ) = 1 u du dx . PROOF d dx (ln | u | ) = d du (ln | u | ) du dx = 1 u du dx . Here are two examples: d dx (ln | 1 x 3 | ) = 1 1 x 3 d dx (1 x 3 ) = 3 x 2 1 x 3 = 3 x 2 x 3 1 . d dx ln x 1 x 2 = d dx (ln | x 1 | ) d dx (ln | x 2 | ) = 1 x 1 1 x 2 . Example 4 Let f ( x ) = x ln x . (a) Specify the domain of f and find the intercepts, if any. (b) On what intervals does f increase? Decrease? (c) Find the extreme values of f . (d) Determine the concavity of the graph and find the points of inflection. (e) Sketch the graph of f . SOLUTION Since the logarithm function is defined only for positive numbers, the domain of f is (0, ) and there is no y -intercept. Since f (1) = 1 · ln 1 = 0, 1 is an x -intercept.
390 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Differentiating f , we have f ( x ) = x · 1 x + ln x = 1 + ln x . To find the critical numbers of f , we set f ( x ) = 0: 1 + ln x = 0, ln x = − 1, x = 1 e (verify this) . Recalling that the logarithm function is increasing on (0, ), and that ln x → −∞ as x 0 + and ln x → ∞ as x → ∞ , we have 1 e 0 f : f ' : 0 1 2 x + + + + + + + + + + + + + + + + + + + + – – – – – – – decreases increases Therefore, f decreases on (0, 1 / e ] and increases on [1 / e , ). By the first derivative test, f (1 / e ) = 1 e ln 1 e = − 1 e = − 0. 368 is a local and absolute minimum of f . Since f ( x ) = 1 / x > 0 for x > 0, the graph of f is concave up on (0, ); there are no points of inflection. You can verify numerically that lim x 0 + x ln x = 0 (see Exercise 30, Section 7.2) and that x ln x → ∞ as x → ∞ . Finally, note that lim x 0 + f ( x ) = lim x 0 + (1 + ln x ) = −∞ .

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