Determine the points of local maxima and local minima:

1. First Derivative Test: If f'(a)= 0 and f'(x) changes the sign of it while passing through point x = a , then

(i) f(x) would have local maximum at x = a if , It means that f'(x) should change it is sign from positive to negative

            For example. f (x) = -x² has local maxima at x = 0.

(ii) f(x) would have local minimum at x = a if . It shows that f'(x) should change it is sign from negative to positive.

            For example. f (x) = x² has local minima at x = 0.

(iii) If f(x) does not change it is sign while passing through x = a, then f (x) would have
 neither a maximum nor minimum at x = a.

            For example f (x) = x³ doesn't have any local maxima or minima at x = 0.

2. Second Derivative Test:

Step I: Let f(x) be differentiable function on given interval and let f'' be continuous at the stationary point. Find f ' (x) and solve equation f ' (x) = 0 given let x = a, b, ... be solutions.

Step II: Case (i) : If f '' (a) < 0 then f(a) is maximum.

             Case (ii): If f ''(a)  > 0 then f(a) is minimum.

Note:

(i) If f''(a) = 0 the 2^nd derivatives test fails in that case we have to go back to the 1^st derivative test.

(ii) If f''(a) = 0 and a is not a point of the local maximum nor local minimum then a is a point of inflection.

3. n^th Derivative Test: It is nothing but general version of the 2^nd derivative test. It says that if,

    f'(a) = f²(a) = f²' (a) =.......... = fⁿ (a) = 0 and fⁿ⁺¹ (a) ≠ 0 (all the derivatives of function up to order 'n' vanishes and (n + 1)th order derivative does not vanish at x = a), then f (x) would have local maximum or minimum at x = a iff n is odd natural number and that x = a would be a point of the local maxima if fⁿ⁺¹ (a) < 0 and would be a point of the local minima if                                  fⁿ⁺¹ (a) > 0.

It is clear that the last 2 tests are basically the mathematical representation of the 1^st derivative test. But that should not diminish the importance of these tests. Because at time it is becomes difficult to decide whether f'(x) changes it is sign or not while passing through point x = a, and remaining tests may come handy in these type of situations.

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