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?
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.
Thank You!
You have correctly found the derivative of the function and simplified it. To find the critical values, you need to set the derivative equal to zero and solve for x. Let's continue from where you left off:
From your work, you have:
(-4 + (x^2)(π/8)(sec(πx/8))^2)/x^2 = 0
Now, to solve this equation, you can multiply both sides by x^2 to get rid of the denominator:
-4 + (x^2)(π/8)(sec(πx/8))^2 = 0
Multiply through by x^2:
-4x^2 + (π/8)(sec(πx/8))^2 = 0
Now, you have a quadratic equation in terms of x. To solve it, you can rearrange the equation:
(π/8)(sec(πx/8))^2 = 4x^2
sec^2(πx/8) = (32πx^2)
Take the square root of both sides:
sec(πx/8) = ±sqrt(32π)x
To find critical values, you need to solve for x. Since sec(πx/8) can never be negative, we consider only the positive square root:
sec(πx/8) = sqrt(32π)x
Now, solving for x directly from this equation is difficult. At this point, it is common to use numerical methods to find the approximate values of the critical points. One such method is the Newton-Raphson method, where you make an initial guess for x and iterate until you find a satisfactory answer.
Alternatively, you can use graphing software or an online plotter to visualize the function and observe the critical points visually.
To summarize, you correctly simplified the derivative and set it equal to zero to find the critical points. However, solving the resulting equation involves solving a trigonometric equation, which can be challenging to do analytically. Using numerical methods or graphical analysis will help you find the approximate values of the critical points.