Engineering Optimization: Theory and Practice, Fourth Edition

66 Classical Optimization Techniques

Theorem 2.2 Sufficient Condition Letf′(x∗)=f′′(x∗) =· · · =f(n−^1 )(x∗) = 0 ,

butf(n)(x∗ )= 0. Thenf (x∗) s (i) a minimum value ofi f (x)iff(n)(x∗) > 0 andn

is even; (ii) a maximum value off (x)iff(n)(x∗) < 0 andnis even; (iii) neither a

maximum nor a minimum ifnis odd.

Proof: Applying Taylor’s theorem with remainder afternterms, we have

f (x∗+ h)=f (x∗) +hf′(x∗)+

h^2

2!

f′′(x∗) +· · · +

hn−^1

(n− 1 )!

f(n−^1 )(x∗)

+

hn

n!

f(n)(x∗+ θh) for 0< θ < 1 (2.5)

Sincef′(x∗)=f′′(x∗) =· · · =f(n−^1 )(x∗) = 0 , Eq. (2.5) becomes

f (x∗+ h)−f (x∗)=

hn

n!

f(n)(x∗+ θh)

Asf(n)(x∗) =0,there exists an interval aroundx∗for every pointxof which thenth

derivativef(n)(x) has the same sign, namely, that off(n)(x∗) Thus for every point.

x∗+ hof this interval,f(n)(x∗+ θh)has the sign off(n)(x∗) When. nis even,hn/n is!

positive irrespective of whetherhis positive or negative, and hencef (x∗+ h)−f (x∗)

will have the same sign as that off(n)(x∗) Thus. x∗will be a relative minimum if

f(n)(x∗) s positive and a relative maximum ifi f(n)(x∗) s negative. Wheni nis odd,

hn/n changes sign with the change in the sign of! hand hence the pointx∗is neither

amaximum nor a minimum. In this case the pointx∗is called a point of inflection.

Example 2.1 Determine the maximum and minimum values of the function

f (x)= 12 x^5 − 54 x^4 + 04 x^3 + 5

SOLUTION Sincef′ (x)= 60 (x^4 − 3 x^3 + 2 x^2 ) = 60 x^2 (x − 1 )(x− 2 ), f′(x) = 0 at

x= 0 , x=1, andx=2. The second derivative is

f′′(x) = 60 ( 4 x^3 − 9 x^2 + 4 x)

Atx= 1 , f′′(x) =−60 and hencex=1 is a relative maximum. Therefore,

fmax= f(x= 1 )= 12

Atx= 2 , f′′(x) = 2 40 and hencex=2 is a relative minimum. Therefore,

fmin= f(x= 2 )= − 11

Atx= 0 , f′′(x) = 0 and hence we must investigate the next derivative:

f′′′(x) = 60 ( 12 x^2 − 81 x+ 4 )=240 at x= 0

Sincef′′′(x) = 0 atx=0,x=0 is neither a maximum nor a minimum, and it is an

inflection point.