What is the extreme f(x)=sqrt(x-2)+sqrt(4-x) ?

The extreme f(x)=sqrt(x-2)+sqrt(4-x) is Minimum(2,sqrt(2)),Maximum(3,2),Minimum(4,sqrt(2))

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extreme f(x)=sqrt(x-2)+sqrt(4-x)

extreme points

f (x) = x - 2 + 4 - x

Minimum (2, 2), Maximum (3, 2), Minimum (4, 2)

Solution steps

f^{'} (x) = \frac{1}{2 x - 2} - \frac{1}{2 4 - x}

Find intervals:

Increasing

: 2 < x < 3,

Decreasing

: 3 < x < 4

Plug

x = 2

into

x - 2 + 4 - x : 2

Minimum (2, 2)

Plug

x = 3

into

x - 2 + 4 - x : 2

Maximum (3, 2)

Plug

x = 4

into

x - 2 + 4 - x : 2

Minimum (4, 2)

Minimum (2, 2), Maximum (3, 2), Minimum (4, 2)

What is the extreme f(x)=sqrt(x-2)+sqrt(4-x) ?
The extreme f(x)=sqrt(x-2)+sqrt(4-x) is Minimum(2,sqrt(2)),Maximum(3,2),Minimum(4,sqrt(2))