Question

state and prove the necessary conditions for the existence of maximum value and minimum value?

Answer 1

It is said that a function has at a point c maximum or minimum if there exists such number $\delta>0$ that for all $x\ \epsilon(c-\delta,c+\delta)$ we have that f(x)$\le f(c)$ or f(x)$\ge\ f(c)$ .

Sufficient Conditions for One Independent Variable

To develop criteria establishing whether a stationary point is a local maximum or minimum, we begin by performing a Taylor series expansion about the stationary point x_o.

y(x) = y(x_o) + y'(x_o) (x - x_o) + ½ y''(x_o) (x - x_o)²+ higher order terms

Now, select x sufficiently close to x_o so the higher order terms become negligible compared to the second-order terms. Since the first derivative is zero at the stationary point, the above equation becomes y(x) = y(x_o) + ½y" (x_o) (x - x_o)

Now we determine if x_o is a local maximum or minimum by examining the value of y"(x_o), since (x - x_o)² is always positive. If y"(x_o) is positive, then the terms ½y"(x_o) (x - x_o)² will always add to y(x_o) in the above equation for x taking on values that are less than or greater than x_o. For this case y(x_o) is a local minimum.

This is summarized in the following:

If y''(x_o) > 0 then y(x_o) is a minimum

y''(x_o) < 0 y(x_o) is a maximum

y''(x_o) = 0 no statement can be made

If the second derivative is zero, it is necessary to examine higher order derivatives. In general if y''(x_o) = ... = y_n-1(x_o) = 0, the Taylor series expansion becomes

y(x) = y(x_o) +$(\frac{1}{n!})$ y(n)(x_o) (x - x_o)ⁿ

If n is even, then (x - x_o)ⁿ is always positive, and the result is:

If y(n)(x_o) > 0 then y(x_o) is a minimum

y(n)(x_o) < 0 y(x_o) is a maximum