What is the extreme f(x)=(x-6)/(x^2-16),0<= x<4 ?

The extreme f(x)=(x-6)/(x^2-16),0<= x<4 is Maximum(0, 3/8),Minimum(2(3-sqrt(5)),(3+sqrt(5))/(16))

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extreme f(x)=(x-6)/(x^2-16),0<= x<4

extreme

Maximum (0, \frac{3}{8}), Minimum (2 (3 - 5), \frac{3 + 5}{16})

Solution steps

Find the critical points:

x = 0, x = 2 (3 - 5)

Domain of

\frac{x - 6}{x ^{2} - 16}, 0 \leq x < 4 : 0 \leq x < 4

Find intervals:

Decreasing

: 0 < x < 2 (3 - 5),

Increasing

: 2 (3 - 5) < x < 4

Plug the extreme point

x = 0

into

\frac{x - 6}{x ^{2} - 16} \Rightarrow y = \frac{3}{8}

Maximum (0, \frac{3}{8})

Plug the extreme point

x = 2 (3 - 5)

into

\frac{x - 6}{x ^{2} - 16} \Rightarrow y = \frac{3 + 5}{16}

Minimum (2 (3 - 5), \frac{3 + 5}{16})

Maximum (0, \frac{3}{8}), Minimum (2 (3 - 5), \frac{3 + 5}{16})

What is the extreme f(x)=(x-6)/(x^2-16),0<= x<4 ?
The extreme f(x)=(x-6)/(x^2-16),0<= x<4 is Maximum(0, 3/8),Minimum(2(3-sqrt(5)),(3+sqrt(5))/(16))