Mathematical Methods for Physics and Engineering - Matematica.NET

2.1 DIFFERENTIATIONgradient of the function is zero but the function rises in the positive x-directionand falls in the negative x-direction is also a stationary point of inflection.The above distinction between the three types of stationary point has beenmade rather descriptively. However, it is possible to define and distinguish stationarypoints mathematically. From their definition as points of zero gradient,all stationary points must be characterised by df/dx = 0. In the case of theminimum, B, the slope, i.e. df/dx, changes from negative at A to positive at Cthrough zero at B. Thus df/dx is increasing and so the second derivative d 2 f/dx 2must be positive. Conversely, at the maximum, Q, we must have that d 2 f/dx 2 isnegative.It is less obvious, but intuitively reasonable, that at S, d 2 f/dx 2 is zero. This maybe inferred from the following observations. To the left of S the curve is concaveupwards so that df/dx is increasing with x and hence d 2 f/dx 2 > 0. To the rightof S, however, the curve is concave downwards so that df/dx is decreasing withx and hence d 2 f/dx 2 < 0.In summary, at a stationary point df/dx = 0 and(i) for a minimum, d 2 f/dx 2 > 0,(ii) for a maximum, d 2 f/dx 2 < 0,(iii) for a stationary point of inflection, d 2 f/dx 2 = 0 and d 2 f/dx 2 changes signthrough the point.In case (iii), a stationary point of inflection, in order that d 2 f/dx 2 changes signthrough the point we normally require d 3 f/dx 3 ≠ 0 at that point. This simplerule can fail for some functions, however, and in general if the first non-vanishingderivative of f(x) at the stationary point is f (n) then if n is even the point is amaximum or minimum and if n is odd the point is a stationary point of inflection.This may be seen from the Taylor expansion (see equation (4.17)) of the functionabout the stationary point, but it is not proved here.◮Find the positions and natures of the stationary points of the functionf(x) =2x 3 − 3x 2 − 36x +2.The first criterion for a stationary point is that df/dx = 0, and hence we setdfdx =6x2 − 6x − 36 = 0,from which we obtain(x − 3)(x +2)=0.Hence the stationary points are at x =3andx = −2. To determine the nature of thestationary point we must evaluate d 2 f/dx 2 :d 2 f=12x − 6.dx2 51