Suppose f is a continuous function defined on a closed interval {a,b}.?

Question

(a) - What theorem guarantees the existence of an absolute maximum value and an absolute minimum value for f?

(b) What steps would you take to find those maximum and minimum values?

Answer 1

a) Be {x(n)} a sequence of numbers in [a, b]


If the maximum does not exist for every M in R there is an x in [a, b] such that f(x)>M. Be {x(n)} a sequence such that for each n f(x(n)) > n

Because the sequence is bound, it contains a sub-sequence that diverges, let l be its limit of that sequence. Because [a, b] is closed, l must be in [a, b]. Because f is continuous the limit of the sub-sequence is f(l) which is a real number.
Thus, the maximum exists, and likewise the minimum exists.

(b) To find the maximum take max {f(a), f(b), the local maximum found}

Answer 2

Mean Value theorem

Take derivative of function and set it equal zero to find any max or min points. Evaluate the function at f(a) and f(b). Check to see which points you have found are the absolute max and min in the closed interval [a,b].

It is assumed that the derivative of the function is continuous throughout the interval [a,b]

Answer 3

the respond isn't conceivable. think f(x) = Y, this is the Y-coordinate. f(2) = -3 corresponds to point A(2, -3) on the coordinate plan. and f(10) = 4 corresponds to point B(10, 4) on the coordinate plan. The question is are you able to draw a line (non-stop line) from factor A to point B with out crossing the X-coordinate. this is not any longer conceivable. the non-stop line you draw is the f(x) function and the fee C is that factor crossing X-coordinate which has Y fee of 0. wish this enables.