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7 INVERSE FUNCTIONS. Certain even and odd combinations of the exponential functions e x and e -x arise so frequently in mathematics and its applications.

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Presentation on theme: "7 INVERSE FUNCTIONS. Certain even and odd combinations of the exponential functions e x and e -x arise so frequently in mathematics and its applications."— Presentation transcript:

1 7 INVERSE FUNCTIONS

2 Certain even and odd combinations of the exponential functions e x and e -x arise so frequently in mathematics and its applications that they deserve to be given special names. INVERSE FUNCTIONS

3 In many ways, they are analogous to the trigonometric functions, and they have the same relationship to the hyperbola that the trigonometric functions have to the circle.  For this reason, they are collectively called hyperbolic functions and individually called hyperbolic sine, hyperbolic cosine, and so on. INVERSE FUNCTIONS

4 7.7 Hyperbolic Functions In this section, we will learn about: Hyperbolic functions and their derivatives. INVERSE FUNCTIONS

5 DEFINITION

6 The graphs of hyperbolic sine and cosine can be sketched using graphical addition, as in these figures. HYPERBOLIC FUNCTIONS

7 Note that sinh has domain and range, whereas cosh has domain and range. HYPERBOLIC FUNCTIONS

8 The graph of tanh is shown.  It has the horizontal asymptotes y = ±1. HYPERBOLIC FUNCTIONS

9 Some mathematical uses of hyperbolic functions will be seen in Chapter 8. Applications to science and engineering occur whenever an entity such as light, Velocity, electricity, or radioactivity is gradually absorbed or extinguished.  The decay can be represented by hyperbolic functions. APPLICATIONS

10 The most famous application is the use of hyperbolic cosine to describe the shape of a hanging wire. APPLICATIONS

11 It can be proved that, if a heavy flexible cable is suspended between two points at the same height, it takes the shape of a curve with equation y = c + a cosh(x/a) called a catenary.  The Latin word catena means ‘chain.’ APPLICATIONS

12 Another application occurs in the description of ocean waves.  The velocity of a water wave with length L moving across a body of water with depth d is modeled by the function where g is the acceleration due to gravity. APPLICATIONS

13 The hyperbolic functions satisfy a number of identities that are similar to well-known trigonometric identities. HYPERBOLIC IDENTITIES

14 We list some identities here.

15 Prove: a.cosh 2 x – sinh 2 x = 1 b.1 – tanh 2 x = sech 2 x HYPERBOLIC FUNCTIONS Example 1

16 HYPERBOLIC FUNCTIONS Example 1 a

17 We start with the identity proved in (a): cosh 2 x – sinh 2 x = 1  If we divide both sides by cosh 2 x, we get: HYPERBOLIC FUNCTIONS Example 1 b

18 The identity proved in Example 1a gives a clue to the reason for the name ‘hyperbolic’ functions, as follows. HYPERBOLIC FUNCTIONS

19 If t is any real number, then the point P(cos t, sin t) lies on the unit circle x 2 + y 2 = 1 because cos 2 t + sin 2 t = 1.  In fact, t can be interpreted as the radian measure of in the figure. HYPERBOLIC FUNCTIONS

20 For this reason, the trigonometric functions are sometimes called circular functions. HYPERBOLIC FUNCTIONS

21 Likewise, if t is any real number, then the point P(cosh t, sinh t) lies on the right branch of the hyperbola x 2 - y 2 = 1 because cosh 2 t - sin 2 t = 1 and cosh t ≥ 1.  This time, t does not represent the measure of an angle. HYPERBOLIC FUNCTIONS

22 However, it turns out that t represents twice the area of the shaded hyperbolic sector in the first figure.  This is just as in the trigonometric case t represents twice the area of the shaded circular sector in the second figure. HYPERBOLIC FUNCTIONS

23 The derivatives of the hyperbolic functions are easily computed.  For example, DERIVATIVES OF HYPERBOLIC FUNCTIONS

24 We list the differentiation formulas for the hyperbolic functions here. DERIVATIVES Table 1

25 DERIVATIVES Note the analogy with the differentiation formulas for trigonometric functions.  However, beware that the signs are different in some cases. Equation 1

26 Any of these differentiation rules can be combined with the Chain Rule.  For instance, Example 2 DERIVATIVES

27 You can see from the figures that sinh and tanh are one-to-one functions.  So, they have inverse functions denoted by sinh -1 and tanh -1. INVERSE HYPERBOLIC FUNCTIONS

28 This figure shows that cosh is not one-to-one. However, when restricted to the domain [0, ∞ ], it becomes one-to-one. INVERSE FUNCTIONS

29 The inverse hyperbolic cosine function is defined as the inverse of this restricted function. INVERSE FUNCTIONS

30 Definition 2 INVERSE FUNCTIONS  The remaining inverse hyperbolic functions are defined similarly.

31 By using these figures, we can sketch the graphs of sinh -1, cosh -1, and tanh -1. INVERSE FUNCTIONS

32 The graphs of sinh -1, cosh -1, and tanh -1 are displayed. INVERSE FUNCTIONS

33 Since the hyperbolic functions are defined in terms of exponential functions, it’s not surprising to learn that the inverse hyperbolic functions can be expressed in terms of logarithms. INVERSE FUNCTIONS

34 In particular, we have: INVERSE FUNCTIONS Eqns. 3, 4, and 5

35 Show that.  Let y = sinh -1 x. Then,  So, e y – 2x – e -y = 0  Or, multiplying by e y, e 2y – 2xe y – 1 = 0  This is really a quadratic equation in e y : (e y ) 2 – 2x(e y ) – 1 = 0 INVERSE FUNCTIONS Example 3

36 Solving by the quadratic formula, we get:  Note that e y > 0, but (because ).  So, the minus sign is inadmissible and we have:  Thus, INVERSE FUNCTIONS Example 3

37 DERIVATIVES Table 6

38 DERIVATIVES Note The formulas for the derivatives of tanh -1 x and coth -1 x appear to be identical. However, the domains of these functions have no numbers in common:  tanh -1 x is defined for | x | < 1.  coth -1 x is defined for | x | >1.

39 The inverse hyperbolic functions are all differentiable because the hyperbolic functions are differentiable.  The formulas in Table 6 can be proved either by the method for inverse functions or by differentiating Formulas 3, 4, and 5. DERIVATIVES

40 Prove that.  Let y = sinh -1 x. Then, sinh y = x.  If we differentiate this equation implicitly with respect to x, we get:  As cosh 2 y - sin 2 y = 1 and cosh y ≥ 0, we have:  So, DERIVATIVES E. g. 4—Solution 1

41 From Equation 3, we have: DERIVATIVES E. g. 4—Solution 2

42 Find.  Using Table 6 and the Chain Rule, we have: DERIVATIVES Example 5

43 Evaluate  Using Table 6 (or Example 4), we know that an antiderivative of is sinh -1 x.  Therefore, DERIVATIVES Example 6


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