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1.5 Exponential and logarithmic functions  (Page 7/17)

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Evaluating hyperbolic functions

  1. Simplify sinh ( 5 ln x ) .
  2. If sinh x = 3 / 4 , find the values of the remaining five hyperbolic functions.
  1. Using the definition of the sinh function, we write
    sinh ( 5 ln x ) = e 5 ln x − e −5 ln x 2 = e ln ( x 5 ) − e ln ( x −5 ) 2 = x 5 − x −5 2 .
  2. Using the identity cosh 2 x − sinh 2 x = 1 , we see that
    cosh 2 x = 1 + ( 3 4 ) 2 = 25 16 .

    Since cosh x ≥ 1 for all x , we must have cosh x = 5 / 4 . Then, using the definitions for the other hyperbolic functions, we conclude that tanh x = 3 / 5 , csch x = 4 / 3 , sech x = 4 / 5 , and coth x = 5 / 3 .
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Simplify cosh ( 2 ln x ) .

( x 2 + x −2 ) / 2

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Inverse hyperbolic functions

From the graphs of the hyperbolic functions, we see that all of them are one-to-one except cosh x and sech x . If we restrict the domains of these two functions to the interval [ 0 , ∞ ) , then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions    . Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.

Definition

Inverse Hyperbolic Functions

sinh −1 x = arcsinh x = ln ( x + x 2 + 1 ) cosh −1 x = arccosh x = ln ( x + x 2 − 1 ) tanh −1 x = arctanh x = 1 2 ln ( 1 + x 1 − x ) coth −1 x = arccot x = 1 2 ln ( x + 1 x − 1 ) sech −1 x = arcsech x = ln ( 1 + 1 − x 2 x ) csch −1 x = arccsch x = ln ( 1 x + 1 + x 2 | x | )

Let’s look at how to derive the first equation. The others follow similarly. Suppose y = sinh −1 x . Then, x = sinh y and, by the definition of the hyperbolic sine function, x = e y − e − y 2 . Therefore,

e y − 2 x − e − y = 0 .

Multiplying this equation by e y , we obtain

e 2 y − 2 x e y − 1 = 0 .

This can be solved like a quadratic equation, with the solution

e y = 2 x ± 4 x 2 + 4 2 = x ± x 2 + 1 .

Since e y > 0 , the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that

y = ln ( x + x 2 + 1 ) .

Evaluating inverse hyperbolic functions

Evaluate each of the following expressions.

sinh −1 ( 2 )
tanh −1 ( 1 / 4 )

sinh −1 ( 2 ) = ln ( 2 + 2 2 + 1 ) = ln ( 2 + 5 ) ≈ 1.4436

tanh −1 ( 1 / 4 ) = 1 2 ln ( 1 + 1 / 4 1 − 1 / 4 ) = 1 2 ln ( 5 / 4 3 / 4 ) = 1 2 ln ( 5 3 ) ≈ 0.2554

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Evaluate tanh −1 ( 1 / 2 ) .

1 2 ln ( 3 ) ≈ 0.5493 .

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Key concepts

  • The exponential function y = b x is increasing if b > 1 and decreasing if 0 < b < 1 . Its domain is ( − ∞ , ∞ ) and its range is ( 0 , ∞ ) .
  • The logarithmic function y = log b ( x ) is the inverse of y = b x . Its domain is ( 0 , ∞ ) and its range is ( − ∞ , ∞ ) .
  • The natural exponential function is y = e x and the natural logarithmic function is y = ln x = log e x .
  • Given an exponential function or logarithmic function in base a , we can make a change of base to convert this function to any base b > 0 , b ≠ 1 . We typically convert to base e .
  • The hyperbolic functions involve combinations of the exponential functions e x and e − x . As a result, the inverse hyperbolic functions involve the natural logarithm.

For the following exercises, evaluate the given exponential functions as indicated, accurate to two significant digits after the decimal.

f ( x ) = 5 x a. x = 3 b. x = 1 2 c. x = 2

a. 125 b. 2.24 c. 9.74

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f ( x ) = ( 0.3 ) x a. x = −1 b. x = 4 c. x = −1.5

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f ( x ) = 10 x a. x = −2 b. x = 4 c. x = 5 3

a. 0.01 b. 10,000 c. 46.42

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f ( x ) = e x a. x = 2 b. x = −3.2 c. x = π

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For the following exercises, match the exponential equation to the correct graph.

  1. y = 4 − x
  2. y = 3 x − 1
  3. y = 2 x + 1
  4. y = ( 1 2 ) x + 4
  5. y = − 3 − x
  6. y = 1 − 5 x
An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -2 to 8. The graph is of a decreasing curved function. The function decreases until it approaches the line “y = 2”, but never touches this line. The y intercept is at the point (0, 3) and there is no x intercept.

d

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An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -9 to 2. The graph is of a function that starts slightly below the line “y = 1” and begins decreasing rapidly in a curve. The x intercept and y intercept are both at the origin.
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An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the x axis and begins increasing rapidly. There is no x intercept and the y intercept is at the point (0, (1/3)). Another point of the graph is at (1, 1).

b

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An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved decreasing function that decreases until it comes close the x axis without touching it. There is no x intercept and the y intercept is at the point (0, 1). Another point of the graph is at (-1, 4).
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An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that increases until it comes close the x axis without touching it. There is no x intercept and the y intercept is at the point (0, -1). Another point of the graph is at (-1, -3).

e

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An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the x axis and begins increasing rapidly. There is no x intercept and the y intercept is at the point (0, 2). Another point of the graph is at (-1, 1).
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For the following exercises, sketch the graph of the exponential function. Determine the domain, range, and horizontal asymptote.

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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