# Showing that the minimum value the continuous function will take too from the extreme value theorem in case M>|m|

From "Extreme value theorem" we say that both maximum and minimum is acheived by the continuous function in its closed interval domain , the proof goes by showing function is bounded and then proved that it needs to be achieved by showing contradiction of letting it not be the case , lets say that upper bound is M so from that proof we can surely say M is acheived , but suppose the maximum and minimum values are not symmetric about y axis , in that case only one is guaranteed to ve achieved from that proof isnt ? What about the other extreme value ? Like for example take this : M (maxima here will be acheived from Boundness theorem) , what about m (M>|m|) (minima how will we show that it will also be achieved ?) Proof which i am referring to Extreme Value Theorem proof help • I don't understand what you are asking. The positions of the max and min have no general relation to x= 0. Apr 23 at 15:26
• x= 0 ? I didnt say anything related to that Apr 23 at 15:41

Since $$f$$ is continuous on $$[a, b]$$, the range of $$f$$ has a lower bound by the boundedness theorem and the greatest lower bound (call it $$m$$) by the completeness of real numbers.
Assume that $$f(x)>m$$ for all $$x\in [a, b]$$. Then function $$g$$ defined on $$[a, b]$$ by $$g(x) = \frac{1}{f(x)-m}$$ is continuous and positive. Hence there is a positive $$K$$ such that $$g(x)\leq K$$ for all $$x\in [a, b]$$ by the boundedness theorem.
Therefore, $$f(x)\geq m+\frac{1}{K}$$ for all $$x\in [a, b]$$. But then $$m+\frac{1}{2K}$$ which is greater than $$m$$ is a lower bound of the range of $$f$$, contradiction the fact that $$m$$ is the greatest lower bound of the range of $$f$$. Therefore, there is a $$c\in [a, b]$$ such that $$f(c)=m$$.