Page 343 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes dy 1
(b) e x e x 0, for maxima or minima
dx 2
–x
x
e = e or x = –x 2x = 0 x = 0.
Second Order Condition:
2
d y 1 x x
e e 1 0 at x = 0 y has a minima at x = 0.
dx 2 2
2
d y 1
(c) Since e x e x 0 for all real values of x, the function has no point of inflexion.
dx 2 2
13.2.1 N Derivative Criterion for Maxima, Minima and Point of Inflexion
th
The criterion for relative maxima or minima of a function y = f(x), discussed so far, fails if f (x) =
0 at the stationary point. Similarly we cannot determine the nature of the point of inflexion if
f (x) = 0 at a point where f (x) = 0. Such situations can be tackled with the help of following n th
derivative criterion.
Let us assume that the first non-zero derivative at a point x = a, encountered in successive
n
derivation, is f (a). Then
n
(i) f(a) will be a maxima if n is even and f (a) < 0.
n
(ii) f(a) will be a minima if n is even and f (a) > 0.
n
(iii) f(a) will be a type I point of inflexion if n is odd and f (a) < 0.
(iv) f(a) will be a type II point of inflexion if n is odd and f (a) > 0.
n

Notes 1. If f(x) has a cusp at x = a, there is either maxima or minima at x = a, although the
above criterion is not applicable.
2. If f(x) has a vertical tangent at x = a, there is a point of inflexion at x = a,
although the above criterion is not applicable.

1
Example: Show that the function y 3 has a point of inflexion at x = 1. What is the
x 1
nature of the point of inflexion?
Solution:

1
y = 3
x 1

dy 3 d y 12
2
= 4 and =
dx x 1 dx 2 x 1 5
2
d y
We note that 2 is not defined at x = 1, therefore, the criterion for point of inflexion is not
dx
applicable.
2
d y
However, since 0 when x < 1 and, > 0 when x > 1, the curve changes from concave to convex
dx 2
and hence the point of inflexion at x = 1 is of type II.

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