# `y = tan((pix)/8), [0,2]` Find the absolute extrema of the function on the closed interval. Given the function `y=tan((pi x)/8)` in [0,2].

We have to find the absolute extrema of the function in the closed interval [0,2].

So first we have to take the derivative of the function and equate it to zero.

i.e `y'=pi/8 sec^2((pi x)/8)=0`

`sec^2((pi x)/8)=0`

Isolating `sec((pi x)/8)=0`

sec(x) cannot be...

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Given the function `y=tan((pi x)/8)` in [0,2].

We have to find the absolute extrema of the function in the closed interval [0,2].

So first we have to take the derivative of the function and equate it to zero.

i.e `y'=pi/8 sec^2((pi x)/8)=0`

`sec^2((pi x)/8)=0`

Isolating `sec((pi x)/8)=0`

sec(x) cannot be equal to 0 because its value ranges between 1 to +infinity and -1 to infinity only.``

So the function does not have a critical point.

So we will calculate the absolute extremas at the end points only.

The absolute minima is y= 0 at x=0 and the absolute maxima is y=1 at x=2.

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