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Published byJoan Reilly Modified over 6 years ago

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**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:

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**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)?

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**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.

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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.

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

Show that

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**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.

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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}.

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**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.

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

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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

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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.

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**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.

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**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,

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