7 INVERSE FUNCTIONS

The Natural Exponential Function INVERSE FUNCTIONS 7.3* The Natural Exponential Function In this section, we will learn about: The natural exponential function and its properties.

THE NATURAL EXPONENTIAL FUNCTION Since ln is an increasing function, it is one-to-one and, therefore, has an inverse function. We denote this by: exp

Thus, according to the definition of an inverse function, NATURAL EXP. FUNCTION Notation 1 Thus, according to the definition of an inverse function,

The cancellation equations are: NATURAL EXP. FUNCTION Equations 2 The cancellation equations are: In particular, we have: exp(0) = 1 since ln 1 = 0 exp(0) = e since ln e = 1

NATURAL EXP. FUNCTION We obtain the graph of y = exp x by reflecting the graph of y = ln x about the line y = x. The domain of exp is the range of ln, that is, (-∞, ∞). The range of exp is the domain of ln, that is, (0, ∞). © Thomson Higher Education

NATURAL EXP. FUNCTION If r is any rational number, the third law of logarithms gives: ln(er ) = r ln e = r Therefore, by Notation 1, exp(r) = er Thus, exp(x) = ex whenever x is a rational number.

NATURAL EXP. FUNCTION This leads us to define ex, even for irrational values of x, by the equation ex = exp(x) In other words, for the reasons given, we define ex to be the inverse of the function ln x.

In this equation, Notation 1 becomes: NATURAL EXP. FUNCTION Notation 3 In this equation, Notation 1 becomes:

The cancellation Equations 2 become: NATURAL EXP. FUNCTION Equations 4 & 5 The cancellation Equations 2 become:

NATURAL EXP. FUNCTION E. g. 1—Solution 1 Find x if ln x = 5. From Notation 3, we see that: ln x = 5 means e5 = x Therefore, x = e5.

Start with the equation ln x = 5 NATURAL EXP. FUNCTION E. g. 1—Solution 2 Start with the equation ln x = 5 Then, apply the exponential function to both sides of the equation: e ln x = e5 However, Equation 4 says that: e ln x = x Therefore, x = e5.

Solve the equation e 5-3x = 10. NATURAL EXP. FUNCTION Example 2 Solve the equation e 5-3x = 10. We take natural logarithms of both sides of the equation and use Equation 5:

NATURAL EXP. FUNCTION Example 2 Since the natural logarithm is found on scientific calculators, we can approximate the solution to four decimal places: x ≈ 0.8991

NATURAL EXP. FUNCTION The exponential function f(x) = ex is one of the most frequently occurring functions in calculus and its applications. So, it is important to be familiar with its graph. © Thomson Higher Education

It is also important to be familiar with its properties. NATURAL EXP. FUNCTION It is also important to be familiar with its properties. These follow from the fact that it is the inverse of the natural logarithmic function.

PROPERTIES Properties 6 The exponential function f(x) = ex is an increasing continuous function with domain and range (0, ∞). Thus, e x > 0 for all x

Also, So, the x-axis is a horizontal asymptote of f(x) = ex. PROPERTIES Properties 6 Also, So, the x-axis is a horizontal asymptote of f(x) = ex.

Find: We divide numerator and denominator by e2x: PROPERTIES Example 3 Find: We divide numerator and denominator by e2x: We have used the fact that t = -2x → -∞ as x → ∞ and so

PROPERTIES Now, we verify that f(x) = ex has the properties expected of an exponential function.

If x and y are real numbers and r is rational, then LAWS OF EXPONENTS Laws 7 If x and y are real numbers and r is rational, then 1. e x+y = e xe y 2. e x-y = e x/e y 3. (e x)r = e rx

Using the first law of logarithms and Equation 5, we have: LAW 1 OF EXPONENTS Proof Using the first law of logarithms and Equation 5, we have: Since ln is a one-to-one function, it follows that: e xe y = e x+y

Laws 2 and 3 are proved similarly. LAWS 2 AND 3 OF EXPONENTS Laws 2 and 3 are proved similarly. See Exercises 95 and 96. As we will see in Section 7.4*, Law 3 actually holds when r is any real number.

DIFFERENTIATION Formula 8 The natural exponential function has the remarkable property that it is its own derivative.

To find its derivative, we use the inverse function method. DIFFERENTIATION Formula 8—Proof The function y = ex is differentiable because it is the inverse function of y = ln x. We know this is differentiable with nonzero derivative. To find its derivative, we use the inverse function method.

DIFFERENTIATION Formula 8—Proof Let y = ex. Then, ln y = x and, differentiating this latter equation implicitly with respect to x, we get:

The geometric interpretation of Formula 8 is: DIFFERENTIATION The geometric interpretation of Formula 8 is: The slope of a tangent line to the curve y = ex at any point is equal to the y-coordinate of the point. © Thomson Higher Education

DIFFERENTIATION This property implies that the exponential curve y = ex grows very rapidly. See Exercise 100. © Thomson Higher Education

Differentiate the function y = e tan x. DIFFERENTIATION Example 4 Differentiate the function y = e tan x. To use the Chain Rule, we let u = tan x. Then, we have y = e u. Hence,

DIFFERENTIATION Formula 9 In general, if we combine Formula 8 with the Chain Rule, as in Example 4, we get:

DIFFERENTIATION Example 5 Find y’ if y = e-4x sin 5x. Using Formula 9 and the Product Rule, we have:

Find the absolute maximum value of the function f(x) = xe -x. DIFFERENTIATION Example 6 Find the absolute maximum value of the function f(x) = xe -x. We differentiate to find any critical numbers: f’(x) = xe-x(-1) + e-x(1) = e-x(1 – x)

Similarly, f ’(x) < 0 when x > 1. DIFFERENTIATION Example 6 Since exponential functions are always positive, we see that f ’(x) > 0 when 1 – x > 0, that is, when x < 1. Similarly, f ’(x) < 0 when x > 1.

DIFFERENTIATION Example 6 By the First Derivative Test for Absolute Extreme Values, f has an absolute maximum value when x = 1. The absolute maximum value is:

DIFFERENTIATION Example 7 Use the first and second derivatives of f(x) = e1/x, together with asymptotes, to sketch its graph. Notice that the domain of f is {x | x ≠ 0}. Hence, we check for vertical asymptotes by computing the left and right limits as x → 0.

As x→ 0+, we know that t = 1/x → ∞. DIFFERENTIATION Example 7 As x→ 0+, we know that t = 1/x → ∞. So, This shows that x = 0 is a vertical asymptote.

As x → 0-, we know that t = 1/x →-∞. So, DIFFERENTIATION Example 7 As x → 0-, we know that t = 1/x →-∞. So, As x→ ∞, we have 1/x → 0. This shows that y = 1 is a horizontal asymptote.

Now, let’s compute the derivative. DIFFERENTIATION Example 7 Now, let’s compute the derivative. The Chain Rule gives: Since e1/x > 0 and x2 > 0 for all x ≠ 0, we have f ’(x) < 0 for all x ≠ 0. Thus, f is decreasing on (-∞, 0) and on (0, ∞). There is no critical number, so the function has no maximum or minimum.

The second derivative is: DIFFERENTIATION Example 7 The second derivative is:

DIFFERENTIATION Example 7 Since e1/x > 0 and x4 > 0, we have f ”(x) > 0 when x > -½(x ≠ 0) and f ”(x) < 0 when x < -½. So, the curve is concave downward on (-∞, -½) and concave upward on (-½, 0) and on (0, ∞). The inflection point is (-½, e-2).

DIFFERENTIATION Example 7 To sketch the graph of f, we first draw the horizontal asymptote y = 1 (as a dashed line) in a preliminary sketch. We also draw the parts of the curve near the asymptotes. © Thomson Higher Education

DIFFERENTIATION Example 7 These parts reflect the information concerning limits and the fact that f is decreasing on both (-∞, 0) and (0, ∞). Notice that we have indicated that f(x) → 0 as x → 0- even though f(0) does not exist. © Thomson Higher Education

DIFFERENTIATION Example 7 We finish the sketch by incorporating the information concerning concavity and the inflection point. © Thomson Higher Education

We check our work with a graphing device. DIFFERENTIATION Example 7 We check our work with a graphing device. © Thomson Higher Education

INTEGRATION Formula 10 As the exponential function y = ex has a simple derivative, its integral is also simple:

Evaluate: We substitute u = x3. Then, du = 3x2 dx. So, and INTEGRATION Example 8 Evaluate: We substitute u = x3. Then, du = 3x2 dx. So, and

Find the area under the curve y = e-3x from 0 to 1. INTEGRATION Example 9 Find the area under the curve y = e-3x from 0 to 1. The area is: