x + x + . Solution We have f ( x ) = x – x + x + f ′ ( x ) = x – x + = ( x – ) f ′ ( x ) = at x = Thus, x = is the only critical point of f . We shall now examine this point for local maxima and/or local minima of f . Observe that f ′ ( x ) ≥ , for all x ∈ R and in particular f ′ ( x ) > , for values close to and to the left and to the right of .
Therefore, by first derivative test, the point x = is neither a point of local maxima nor a point of local minima. Hence x = is a point of inflexion. Remark One may note that since f ′ ( x ), in Example , never changes its sign on R , graph of f has no turning points and hence no point of local maxima or local minima. We shall now give another test to examine local maxima and local minima of a given function.
This test is often easier to apply than the first derivative test. Theorem (Second Derivative Test) Let f be a function defined on an interval I and c ∈ I. Let f be twice differentiable at c . Then (i) x = c is a point of local maxima if f ′ ( c ) = and f ″ ( c ) < The value f ( c ) is local maximum value of f .
(ii) x = c is a point of local minima if c = ′ and f ″ ( c ) > In this case, f ( c ) is local minimum value of f . (iii) The test fails if f ′ ( c ) = and f ″ ( c ) = . In this case, we