- #1

fishspawned

- 66

- 15

## Homework Statement

i am looking for the potential extrema on the function y = |sinx(x)| + (1/x) between [-2pi, 2pi]

## Homework Equations

extrema will be likely located at f'(x) = 0 or u/d

## The Attempt at a Solution

first it is noted that there is a discontinuity at x=0

then determining f'(x) by using |sin(x)| as sqrt(sin

^{2}x) instead

we get f'(x) = sin(x)cos(x)/|sinx(x)| - x

^{-2}

first the easy one : we can see that f'(x) = u/d where sin(x) = 0 [looking only at the denominator here]

that will be at x= -2pi, -pi, pi, and 2pi

My problem is the second part

f'(x) = 0 when the whole equation = 0

0 = sin(x)cos(x)/|sinx(x)| - x

^{-2}

therefore

sin(x)cos(x)/|sinx(x)| = x

^{-2}

using |sin(x)| as sqrt(sin

^{2}x) instead

we get

sin(x)cos(x)/(sqrt(sin

^{2}x)) = x

^{-2}

a little algebra and then we get

x

^{4}= sec

^{2}(x)

and now.... i am stuck.

is there a way to solve this particular equation?