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Unit 13: Maxima and Minima

13.1.2 Second Derivative Criterion for Local Extrema Notes

When the function f(x) is twice differentiable at an interior point c of the domain, then
(i) f(x) has a local maxima at x = c if f (c) = 0 and f (c) < 0.
(ii) f(x) has a local minima at x = c if f (c) = 0 and f (c) > 0.

Notes When f(x) has a maxima (or minima) at c, the curve of f(x) is concave (or convex)
from below. This test is inconclusive when f (c) = 0.

Example:
(a) Show that the function y = x – 2x + 3 has a minima at x = 1. Find the minimum value of the
2
function.
2
(b) Show that the function y = x – 2x + 3 has a maxima at x = 5/2. Find the maximum value of y.
Solution:

dy
2
(a) We have y = y = x – 2x + 3 = 2x – 2 = 0, for maxima or minima.
dx
dy
2(x –1) = 0 or x = 1 is a stationary point (A point at which 0 ).
dx
To know whether y is maximum or minimum at x = 1, we determine the sign of second
derivative at this point.
2
d y
Since 2 0, therefore the function has a minima at x = 1.
dx 2
Further, the minimum value of y = 1 – 2 + 3 = 2.
2
dy
(b) We have y = 100 + 15x – 3x 2 = 15 – 6x = 0, for maxima or minima. This implies that
dx
15 5
x is a stationary point.
6 2
2
d y 5
Since = –6 < 0, therefore, the function has a maxima at x . The maximum value of
dx 2 2
15 5 3 25
the function is given by y 100 118.75.
2 4
Procedure for fuiding absolute extremia

To find absolute extrema of a continuous function f(x) on [a, b]

1. Find all critical points of f(x) on [a, b].
2. Evaluate f(x) at each critical points as well as at the end points a and b.
3. The largest-value of f(x), obtained above, is absolute maxima and the smallest-value is
absolute minima.

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