What is the extreme f(x)=(x^2+81)/(4x),1<= x<= 12 ?

The extreme f(x)=(x^2+81)/(4x),1<= x<= 12 is Maximum(1, 41/2),Minimum(9, 9/2),Maximum(12, 75/16)

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extreme f(x)=(x^2+81)/(4x),1<= x<= 12

extreme points

f (x) = \frac{x ^{2} + 81}{4 x}, 1 \leq x \leq 12

Maximum (1, \frac{41}{2}), Minimum (9, \frac{9}{2}), Maximum (12, \frac{75}{16})

Solution steps

Find the critical points:

x = 1, x = 9, x = 12

Domain of

\frac{x ^{2} + 81}{4 x}, 1 \leq x \leq 12 : 1 \leq x \leq 12

Find intervals:

Decreasing

: 1 < x < 9,

Increasing

: 9 < x < 12

Plug the extreme point

x = 1

into

\frac{x ^{2} + 81}{4 x} \Rightarrow y = \frac{41}{2}

Maximum (1, \frac{41}{2})

Plug the extreme point

x = 9

into

\frac{x ^{2} + 81}{4 x} \Rightarrow y = \frac{9}{2}

Minimum (9, \frac{9}{2})

Plug the extreme point

x = 12

into

\frac{x ^{2} + 81}{4 x} \Rightarrow y = \frac{75}{16}

Maximum (12, \frac{75}{16})

Maximum (1, \frac{41}{2}), Minimum (9, \frac{9}{2}), Maximum (12, \frac{75}{16})

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

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 5 x + 2

f (x) = x e^{- \frac{x ^{2}}{32}} [- 3.8]

U (X, Y) = 2 X + 5 Y

f (x, y) = 5 x^{2} - 5 y^{2}

What is the extreme f(x)=(x^2+81)/(4x),1<= x<= 12 ?
The extreme f(x)=(x^2+81)/(4x),1<= x<= 12 is Maximum(1, 41/2),Minimum(9, 9/2),Maximum(12, 75/16)