Download presentation

Published byJoan Reilly Modified over 2 years ago

2
**The inverse of f (x), denoted f −1(x), is the function that reverses the effect of f (x).**

For example, the inverse of f (x) = x3 is the cube root function f −1(x) = x1/3. Given a table of function values for f (x), we obtain a table for f −1(x) by interchanging the x and y columns:

3
**DEFINITION Inverse Let f (x) have domain D and range R**

DEFINITION Inverse Let f (x) have domain D and range R. If there is a function g (x) with domain R such that then f (x) is said to be invertible. The function g (x) is called the inverse function and is denoted f −1(x). Show that f (x) = 2x − 18 is invertible. What are the domain and range of f −1(x)?

4
**So when does a function f (x) have an inverse**

So when does a function f (x) have an inverse? The answer is: When f (x) is one-to-one, which means that f (x) takes on each value in its range at most once.

5
When f (x) is one-to-one on its domain D, the inverse function f −1(x) exists and its domain is equal to the range R of f. Indeed, for every c R, there is precisely one element a D such that f (a) = c and we may define f −1(c) = a. With this definition, f (f −1(c)) = f (a) = c and f −1(f (a)) = f −1(c) = a. This proves the following theorem. THEOREM 1 Existence of Inverses The inverse function f −1(x) exists if and only if f (x) is one-to-one on its domain D. Furthermore, Domain of f = range of f −1. Range of f = domain of f −1.

6
**is invertible. Find f −1 & determine the domain and range of f and f −1.**

Show that

7
**Show that f (x) = x5 + 4x + 3 is one-to-one.**

Often, it is impossible to find a formula for the inverse because we cannot solve for x explicitly in the equation y = f (x). For example, the function f(x) = x + ex has an inverse, but we must make do without an explicit formula for it. Show that f (x) = x5 + 4x + 3 is one-to-one. If n odd and c > 0, then cxn is increasing. A sum of increasing functions is increasing. The increasing function f (x) = x5 + 4x + 3 satisfies the Horizontal Line Test.

8
Restricting the Domain Find a domain on which f (x) = x is one-to-one and determine its inverse on this domain. f (x) = x2 satisfies the Horizontal Line Test on the domain {x : x ≥ 0}.

9
**Next we describe the graph of the inverse function**

Next we describe the graph of the inverse function. The reflection of a point (a, b) through the line y = x is, by definition, the point (b, a). Note that if the x-and y-axes are drawn to the same scale, then (a, b) and (b, a) are equidistant from the line y = x and the segment joining them is perpendicular to y = x. The reflection (a, b) through the line y = x is the point (b, a). The graph of f −1(x) is the reflection of the graph of f (x) through the line y = x.

10
**Sketching the Graph of the Inverse Sketch the graph of the inverse of**

11
THEOREM 2 Derivative of the Inverse Assume that f (x) is differentiable and one-to-one with inverse g(x) = f −1(x). If b belongs to the domain of g(x) and f (g (b)) 0, then g (b) exists and

12
GRAPHICAL INSIGHT The formula for the derivative of the inverse function has a clear graphical interpretation. Consider a line L of slope m and let L be its reflection through y = x. Then the slope of L is 1/m. Indeed, if (a, b) and (c, d) are any two points on L, then (b, a) and (d, c) lie on L and Now recall that the graph of the inverse g (x) is obtained by reflecting the graph of f (x) through the line y = x. As we can see, the tangent line to y = g (x) at x = b is the reflection of the tangent line to y = f (x) at x = a [where b = f (a) and a = g (b)]. These tangent lines have reciprocal slopes, and thus g (b) = 1/f (a) = 1/f (g (b)), as claimed in Theorem 2.

13
**Calculate g (x), where g(x) is the inverse of the function**

f (x) = x on the domain {x : x ≥ 0}. We obtain this same result by differentiating g (x) = (x − 10)1/4 directly.

14
**Calculate g (1), where g (x) is the inverse of f (x) = x + ex.**

In this case, we cannot solve for g (x) explicitly, but a formula for g (x) is not needed. All we need is the particular value g (1), which we can find by solving f (x) = 1. By inspection, x + ex = 1 has solution x = 0. Therefore, f (0) = 1 and, by definition of the inverse, g (1) = 0. Since f (x) = 1 + ex,

Similar presentations

OK

Goal: Find and use inverses of linear and nonlinear functions.

Goal: Find and use inverses of linear and nonlinear functions.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on farm management Biology ppt on reproduction Ppt on teacher's day Ppt on games and sports in india Ppt on different solid figures chart Ppt on centring sheet Ppt on media and entertainment industry Ppt on conservation and management of soil Ppt on charge coupled device for side Ppt on central administrative tribunal delhi