Question

How do I find the critical values?

y= 4/x + tan(πx/8)

What I did is I simplified it to

y= 4x^-1 + tan(πx/8)

then I took the derivative

y'= -4x^-2 + (π/8)(sec(πx/8))^2

Then I simplied it

y'= -4/x^2 + (π/8)(sec(πx/8))^2

then I found a common denomiator and combined the two values

Y'= (-4 + (x^2)(π/8)(sec(πx/8))^2)/x^2

Then I set the top and the bottom equal to zero

x^2 = 0

x=0

I got this one but with the other one

-4 + (x^2)(π/8)(sec(πx/8))^2 = 0

(x^2)(π/8)(sec(πx/8))^2 = 4

(x^2)(sec(πx/8))^2 = 32/π

but now I don't know what to do, did I do everything right up to now and how to I solve from here?

Answer 1

Nasty nasty question, no wonder you messed up

picking it up at

y'= -4/x^2 + (π/8)(sec(πx/8))^2

= -32/(8x^2) + (πx^2/8x^2) sec^2 (πx/8)
= (-1/8x^2) (32 - π x^2 sec^2 (πx/8))
= 0 for critical values

so -1/8x^2) = 0 ---> no solution
or
32 - π x^2 sec^2 (πx/8) = 0

here is where it gets interesting.

x^2 sec^2 (πx/8) = 32/π

YUP, you had that, good for you

let's see what Wolfram has to say about that.

http://www.wolframalpha.com/input/?i=solve+x%5E2+sec%5E2+%28πx%2F8%29+%3D+32%2Fπ
it says x = appr ± 2.134

There is no nice way to solve this type of equation.
The link I gave you is one of the best there is to solve this type of problem.

Answer 2

Thank You!