2.3 The Chain Rule and Inverse Functions

Transcription

1 2.3 The Chain Rule and Inverse Functions 2.3. The Chain Rule and its Applications Theorem 2.3. (Chain Rule). If f(x) and g(x) are differentiable, then the derivative of the composite function (f g)(x) =f(g(x)) is given by (f g) 0 (x) =f 0 (g(x))g 0 (x), thatis, d dx [f(g(x))] = f 0 (g(x))g 0 (x). The derivative of the outside (leave the inside alone) times the derivative of the inside. Theorem (General Power Rule). If g(x) is differentiable, then the derivative of the composite function (g(x)) n, where n is a real number, is given by d dx [(g(x))n ]=n(g(x)) n g 0 (x). Problem 4. Find the derivative of the following functions. h(x) =(x 3 +3) h(x) =0(x 2 + x) h(x) =3(x 3 x 2 +4) 7/3. 4. h(x) = p x 3 x 2 +4x h(x) = (x 2 +3) h(x) = x3 +3 (x 2 +3) h(x) =(x 3 +3) 3 (x 2 +3) 5. Definition We write f : I! R to mean that f is a function from the set (or interval) I to the set R, thatis,foreveryelementofx 2 I there is a unique element f(x) in R. The value f(x) is called. f of x, or 2. the image of x under f. We also say that the element x 2 I gets mapped to the element f(x) by the function f. The set I is called the domain of the function f and the set R is called the co-domain of the function f. Definition Given a function f : I! R the range of f, denotedbyran(f), istheset ran(f) ={f(x) :x 2 I} = {y 2 R : y = f(x) for some x 2 I}. Example. Apply the above definition and address the following:. Using the function f : I! R in Figure 29, determine the range of f. 2. Let f : R! R be defined by f(x) =x 2 x.whatisf(3)? Whatistherangeoff. 63

2 Figure 29: Graph of a function f : I! R One-To-One Functions Definition Afunctionf : I! R is said to be one-to-one (or an injection), if distinct elements in I get mapped to distinct elements in R; morespecifically or equivalently, for all x 2 I and x 2 2 I, ifx 6= x 2,thenf(x ) 6= f(x 2 ) for all x 2 I and x 2 2 I, iff(x )=f(x 2 ),thenx = x 2. Problem 5. Below are the graphs of a function g : I! R and a function f : J! R. Is g one-to-one? Is f one-to-one? g: I! R f : J! R Horizontal Line Test. Afunctionisone-to-oneifnohorizontallineintersectsitsgraph more than once. Recall the following definition discussed in Chapter. Definition Let f : I! R be a function where I is an interval. We say that f is strictly increasing on I when the following holds: For all x,x 2 2 I if x <x 2,thenf(x ) <f(x 2 ). We say that f is strictly decreasing on I when the following holds: For all x,x 2 2 I if x <x 2,thenf(x ) >f(x 2 ). Theorem (One-to-One Test). Let f : I! R be a function where I is an interval. If f is a strictly increasing or a strictly decreasing, then f is one-to-one on the interval I. 64

3 Corollary (Derivative Test). Suppose that f : I! R is a differentiable function. If f 0 (x) > 0 for all x in I, thenf is one-to-one. If f 0 (x) < 0 for all x n I, thenf is one-to-one. Problem 6. Let f : R! R be defined by f(x) =x 5 +2x +.Showthatf is one-to-one Inverse Functions Definition Afunctionf has an inverse function, denoted by f,if. f(f (x)) = x for all x in the domain of f 2. f (f(x)) = x for all x in the domain of f. The domain of f is the range of f. Anotherwaytosaythatf is the inverse of f is when for all real numbers a and b, f (a) =b if and only if f(b) =a. (N) Remark Given a calculus function f(x) which is one-to-one. To find a formula for f (y) try the following steps:. Write the equation y = f(x). 2. Solve the equation y = f(x) for the unique x, obtainingx = g(y). 3. In g(y) replace the variable y with x. 4. Then f (x) =g(x). Example. One can prove that the function f : R! R defined by f(x) =(x one-to-one. Find a formula for the inverse function f. 4) 3 +2 is Questions. Given a function f, then one can ask the following:. When does f exist? 2. What is the graph of f? 3. Is it always possible to get a formula for f? 4. What is the derivative of f? Answers.. f exists whenever f is one-to-one. 2. The graph of f is the reflection of the graph of f(x) about the line y = x (see the example below). 3. In most cases it is impossible to get a formula for f. 4. See Theorem 2.3. below. 65

4 Example. The figure below contains the graph of f(x) = x3 + x +2. Notice that the function f(x) is one-to-one and therefore, has an inverse function f (x). The graph of f (x) is obtained by reflecting the graph of f(x) about the line y = x. Thus, the graph of f (x) is also given in the figure. y 8 6 y f x y x 4 2 y f x 0 x Problem 7. Consider the one-to-one function f(x) = x3 8 + x Evaluate f (2) and f (4). Solution. By (N) ofdefinition2.3.9forallrealnumbersa and b, wehavethatf (a) =b if and only if f(b) =a. To evaluate f (2) we note that f (2) = x if and only if f(x) =2. Thus, solving the equation x3 8 + x 2 +2=2for x gives x =0. Therefore, f (2) = 0. To evaluate f (4) we note that f (4) = x if and only if f(x) =4. Thus, solving the equation x3 8 + x 2 +2=4for x gives x =2. Therefore, f (4) = 2. Remark. The function f(x) = x3 find a formula for f (x). 8 + x The Inverse Function Theorem +2is one-to-one. It is algebraically very difficult to Theorem 2.3. (Inverse Function Theorem). Let f : I! R be differentiable on the interval I. LetJ be the range of f. Supposethatf 0 (x) 6= 0for all x 2 I. Then f is one-to-one, J is an interval, and f : J! I is differentiable on J. Moreover, for all x 2 J. (f ) 0 (x) = f 0 (f (x)), Proof. To see why f is differentiable at x, see Theorems 2.A.2 and 2.A.3 in the text. We shall now derive the above formula. Since f(f (x)) = x for all x, we shall differentiate both sides of this equation. Using the chain rule on the left side, we obtain f 0 (f (x)) (f ) 0 (x) =. 66

5 Thus, solving for (f ) 0 (x) we obtain This completes the proof. (f ) 0 (x) = f 0 (f (x)). Problem 8. Let f : R! R be defined by f(x) =x 5 +2x +. By Problem 6, we know that f is one-to-one and hence, the inverse function f exists. Evaluate (f ) 0 (4). Solution. We cannot find a formula for f (x). So, we shall apply Theorem First we find an x satisfying f(x) =4,thatis,x 5 +2x +=4. Clearly, x =satisfies f() = 4. Thus, f (4) =. Theorem 2.3. now implies that (f ) 0 (4) = f 0 (f (4)) = f 0 (). Since f 0 (x) =5x 4 +2,weseethatf 0 () = = 7. Therefore, (f ) 0 (4) = f 0 (f (4)) = f 0 () = 7. Problem 9. Let f : R! R be defined by f(x) = x3 + x +2.Weknowthatf is one-to-one and hence, the inverse function f exists. Evaluate (f ) 0 (4). Solution. First we find an x satisfying f(x) =4,thatis, x3 + x +2 = 4. Clearly, x =2 satisfies f(2) = 4. Thus, f (4) = 2. Theorem 2.3. now implies that (f ) 0 (4) = f 0 (f (4)) = f 0 (2). Since f 0 (x) = 3x ,weseethatf 0 (2) = 2. Therefore, (f ) 0 (4) = f 0 (f (4)) = f 0 (2) = 2. Problem 0. Let f : R! R be defined by f(x) = x3 + x +2.Weknowthatf is one-to-one and hence, the inverse function f exists. Evaluate (f ) 0 (0). Solution. First we find an x satisfying f(x) =0,thatis, x3 + x +2=0. Clearly, x = 2 satisfies f( 2) = 0. Thus, f (0) = 2. Theorem 2.3. now implies that (f ) 0 (0) = f 0 (f (0)) = f 0 ( 2). Since f 0 (x) = 3x ,weseethatf 0 ( 2) = 2. Therefore, (f ) 0 (0) = f 0 (f (0)) = f 0 ( 2) = 2. 67