What is the extreme f(x)=y=tan((pix)/8),0<= x<= 2 ?

The extreme f(x)=y=tan((pix)/8),0<= x<= 2 is Minimum(0,0),Maximum(2,1)

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extreme f(x)=y=tan((pix)/8),0<= x<= 2

extreme points

f (x) = y = tan (\frac{π x}{8}), 0 \leq x \leq 2

Minimum (0, 0), Maximum (2, 1)

Solution steps

Find the critical points:

x = 0, x = 2

Domain of

tan (\frac{π x}{8}), 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Increasing

: 0 < x < 2

Plug the extreme point

x = 0

into

tan (\frac{π x}{8}) \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 2

into

tan (\frac{π x}{8}) \Rightarrow y = 1

Maximum (2, 1)

Minimum (0, 0), Maximum (2, 1)

x^{4} - y^{3} - 2 x^{2} + 3 y + 1

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

f (x, y) = \frac{1}{x - y - 4}

f (x) = \frac{x ^{3}}{x ^{2} - 36}

f (x) = x^{3} - x^{4}

What is the extreme f(x)=y=tan((pix)/8),0<= x<= 2 ?
The extreme f(x)=y=tan((pix)/8),0<= x<= 2 is Minimum(0,0),Maximum(2,1)