Amenability by Alan L. T. Paterson PDF

By Alan L. T. Paterson

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Hence the slope of the tangent is decreasing. This means that the second derivative must be negative. Chapter 3 : Introduction to Derivatives page 8 of 20 c First Year Calculus W W L Chen, 1982, 2008 The above heuristics can be summarized by the following result on stationary points and second derivatives which we shall establish formally in Chapter 8. PROPOSITION 3E. Suppose that I is an open interval containing a real number a. Suppose further that the function f (x) is differentiable at every x ∈ I, and that f (a) = 0.

Consider next the function f (x) = x3 . At any given point x, let us consider a small increment ∆x and the behaviour of the function as x changes to x + ∆x. Clearly the value f (x) changes to f (x + ∆x), giving rise to the error ∆f = f (x + ∆x) − f (x) = (x + ∆x)3 − x3 = 3x2 ∆x + 3x(∆x)2 + (∆x)3 , and the relative error f (x + ∆x) − f (x) ∆f = = 3x2 + 3x∆x + (∆x)2 . ∆x ∆x As ∆x is taken to be very small, we have respectively the approximations ∆f ≈ 3x2 ∆x and ∆f ≈ 3x2 . ∆x Note again that the first of these suggests that ∆f is essentially directly proportional to ∆x, and the second shows that the relative error is an approximation of the derivative.

STATIONARY POINTS. We now use our knowledge on derivatives to further our understanding of the functions. 2. 13. Let us continue our investigation of the function f (x) = 5 + x−3 . Simple calculation gives f (x) = −3x−4 . It follows that there is no stationary point. Next, note that f (x) < 0 whenever x = 0. It follows that the function is decreasing in the open intervals (−∞, 0) and (0, +∞). We now supplement our earlier effort with this extra information. 14. Let us continue our investigation of the function f (x) = 1/(x − 1)(x − 2).

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