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

Textbook Question

Chapter 3, 3.1 - Problem 36 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

1 Answer | Add Yours

nees101's profile pic

nees101 | Student, Graduate | (Level 2) Adjunct Educator

Posted on

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.

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question