What is the extreme f(x)=sqrt(12+x)+sqrt(24-x) ?

The extreme f(x)=sqrt(12+x)+sqrt(24-x) is Minimum(-12,6),Maximum(6,6sqrt(2)),Minimum(24,6)

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extreme f(x)=sqrt(12+x)+sqrt(24-x)

extreme points

f (x) = 12 + x + 24 - x

Minimum (- 12, 6), Maximum (6, 62), Minimum (24, 6)

Solution steps

f^{'} (x) = \frac{1}{2 12 + x} - \frac{1}{2 24 - x}

Find intervals:

Increasing

: - 12 < x < 6,

Decreasing

: 6 < x < 24

Plug

x = - 12

into

12 + x + 24 - x : 6

Minimum (- 12, 6)

Plug

x = 6

into

12 + x + 24 - x : 62

Maximum (6, 62)

Plug

x = 24

into

12 + x + 24 - x : 6

Minimum (24, 6)

Minimum (- 12, 6), Maximum (6, 62), Minimum (24, 6)

f (x) = - \frac{1}{12} x^{3} + x - \frac{1}{3}

f (x, y) = 2 x^{2} + y^{2} - y

f (x, y) = \frac{1}{2} x^{2} - x y + \frac{1}{3} y^{3}

f (x) = \frac{- ( \frac{1}{2} ) x ^{4} + 2}{3 x ^{3} - 4}

\frac{( x - 2 ) ^{3}}{( x - 4 ) ^{4}}

What is the extreme f(x)=sqrt(12+x)+sqrt(24-x) ?
The extreme f(x)=sqrt(12+x)+sqrt(24-x) is Minimum(-12,6),Maximum(6,6sqrt(2)),Minimum(24,6)