extreme value theorem | ab calculus bc,12th grade math

# Extreme Value Theorem

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Extreme value theorem (EVT) states that if the function 'f' is continuous on a closed interval [a,b] then the function 'f' has minimum and maximum on the interval.
$\blacktriangleright$ If the interval is not closed then the EVT can fail.
$\blacktriangleright$ If the function is not continuous then the EVT can fail.
Explanation of the above graph are as follows :
Fig.(a) Maximum value at f(2)= 4 and minimum at f(4.3)= 2
Fig.(b) Maximum value at f(3.1)= 4 and minimum at f(5)= 1
Fig.(c) Sometimes there are more than one maximum or more than one minimum or no maximum or no minimum at all. Maximum value at f(2.1)= 6 f(5)= 6 and minimum at f(4)= 2
Fig.(d) No maximum and minimum at f(5)= 0
Fig.(e) No maximum and no minimum

## Examples on Extreme Value Theorem

Check whether the in the following questions extreme value theorem is applicable or not.
1) f(x) = -$x^{3} -6x^{2}$ - 9x in the closed interval [-3,-1].
Solution : As the given function is a cubic polynomial and all the polynomials are continuous and the interval given is a closed interval so the EVT theorem is applicable. So the given function has maximum or minimum values.

2) f(x) = $\frac{x^{2}-4}{x-2}$ in the closed interval [-1,5].
Solution : As the given function has a hole at x = 2 that means the function is discontinuous in the closed interval so EVT theorem is not applicable. So no maximum or minimum values.

3) f(x) = $\sqrt{4-x^{2}}$ in the closed interval [-2,2].
Solution : When we graph the given function using graphing calculator or in desmos you will get to know that the given function is continuous in the closed interval [-2,2] so EVT theorem is applicable. So the given function has maximum or minimum values. Check the graph below.