1.5 Exponential and logarithmic functions (Page 7/17)

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Evaluating hyperbolic functions

Simplify $sinh (5 ln x) .$
If $sinh x = 3 / 4,$ find the values of the remaining five hyperbolic functions.

Using the definition of the $sinh$ function, we write
$sinh (5 ln x) = \frac{e^{5 ln x} - e^{−5 ln x}}{2} = \frac{e^{ln (x^{5})} - e^{ln (x^{−5})}}{2} = \frac{x^{5} - x^{−5}}{2} .$
Using the identity ${cosh}^{2} x - {sinh}^{2} x = 1,$ we see that
${cosh}^{2} x = 1 + {(\frac{3}{4})}^{2} = \frac{25}{16} .$

Since $cosh x \geq 1$ 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 $[0, \infty),$ 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

\begin{matrix} {sinh}^{−1} x = arcsinh x = ln (x + \sqrt{x^{2} + 1}) & {cosh}^{−1} x = arccosh x = ln (x + \sqrt{x^{2} - 1}) \\ {tanh}^{−1} x = arctanh x = \frac{1}{2} ln (\frac{1 + x}{1 - x}) & {coth}^{−1} x = arccot x = \frac{1}{2} ln (\frac{x + 1}{x - 1}) \\ {sech}^{−1} x = arcsech x = ln (\frac{1 + \sqrt{1 - x^{2}}}{x}) & {csch}^{−1} x = arccsch x = ln (\frac{1}{x} + \frac{\sqrt{1 + x^{2}}}{| x |}) \end{matrix}

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, $x = \frac{e^{y} - e^{− y}}{2} .$ 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} = \frac{2 x \pm \sqrt{4 x^{2} + 4}}{2} = x \pm \sqrt{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 + \sqrt{x^{2} + 1}) .

Evaluating inverse hyperbolic functions

Evaluate each of the following expressions.

{sinh}^{−1} (2)

{tanh}^{−1} (1 / 4)

${sinh}^{−1} (2) = ln (2 + \sqrt{2^{2} + 1}) = ln (2 + \sqrt{5}) \approx 1.4436$

${tanh}^{−1} (1 / 4) = \frac{1}{2} ln (\frac{1 + 1 / 4}{1 - 1 / 4}) = \frac{1}{2} ln (\frac{5 / 4}{3 / 4}) = \frac{1}{2} ln (\frac{5}{3}) \approx 0.2554$

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Evaluate ${tanh}^{−1} (1 / 2) .$

$\frac{1}{2} ln (3) \approx 0.5493 .$

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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 $(− \infty, \infty)$ and its range is $(0, \infty) .$
The logarithmic function $y = {log}_{b} (x)$ is the inverse of $y = b^{x} .$ Its domain is $(0, \infty)$ and its range is $(− \infty, \infty) .$
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 $b > 0, b \neq 1 .$ 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. $x = \frac{1}{2}$ c. $x = \sqrt{2}$

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. $x = \frac{5}{3}$

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.

$y = 4^{− x}$
$y = 3^{x - 1}$
$y = 2^{x + 1}$
$y = {(\frac{1}{2})}^{x} + 4$
$y = − 3^{− x}$
$y = 1 - 5^{x}$

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.

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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).

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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).

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For the following exercises, sketch the graph of the exponential function. Determine the domain, range, and horizontal asymptote.

Questions & Answers

find the equation of the tangent to the curve y=2x³-x²+3x+1 at the points x=1 and x=3

Esther Reply

derivative of logarithms function

Iqra Reply

how to solve this question

sidra

ex 2.1 question no 11

khansa

anyone can help me

khansa

question please

Rasul

ex 2.1 question no. 11

khansa

i cant type here

khansa

Find the derivative of g(x)=−3.

Abdullah Reply

any genius online ? I need help!!

Guzorochi Reply

how can i help you?

Pina

need to learn polynomial

Zakariya

i will teach...

nandu

I'm waiting

Zakariya

plz help me in question

Abish

How can I help you?

Tlou

evaluate the following computation (x³-8/x-2)

Murtala Reply

teach me how to solve the first law of calculus.

Uncle Reply

teach me also how to solve the first law of calculus

Bilson

what is differentiation

Ibrahim Reply

only god knows😂

abdulkadir

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)

Naufal Reply

fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5

diron

pweding paturo nsa calculus?

jimmy

how to use fundamental theorem to solve exponential

JULIA Reply

find the bounded area of the parabola y^2=4x and y=16x

Omar Reply

what is absolute value means?

Geo Reply

Chicken nuggets

Hugh

🐔

🐔🦃 nuggets

(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.

Ismael

find integration of loge x

Game Reply

find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3

Godwin Reply

how does this work

Brad Reply

Can calculus give the answers as same as other methods give in basic classes while solving the numericals?

Cosmos Reply

log tan (x/4+x/2)

Rohan

please answer

Rohan

y=(x^2 + 3x).(eipix)

Claudia

is this a answer

Ismael

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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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