What is the extreme f(x)=tan((pix)/(32)),0<= x<= 8 ?

The extreme f(x)=tan((pix)/(32)),0<= x<= 8 is Minimum(0,0),Maximum(8,1)

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extreme f(x)=tan((pix)/(32)),0<= x<= 8

extreme

Minimum (0, 0), Maximum (8, 1)

Solution steps

Find the critical points:

x = 0, x = 8

Domain of

tan (\frac{π x}{32}), 0 \leq x \leq 8 : 0 \leq x \leq 8

Find intervals:

Increasing

: 0 < x < 8

Plug the extreme point

x = 0

into

tan (\frac{π x}{32}) \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 8

into

tan (\frac{π x}{32}) \Rightarrow y = 1

Maximum (8, 1)

Minimum (0, 0), Maximum (8, 1)

What is the extreme f(x)=tan((pix)/(32)),0<= x<= 8 ?
The extreme f(x)=tan((pix)/(32)),0<= x<= 8 is Minimum(0,0),Maximum(8,1)