What is the relative maximum of y = csc (x)?

Question

13059 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-09T23:16:12

$y = csc x = \frac{1}{sin x} = {(sin x)}^{- 1}$ $y=cscx=1/sinx=(sinx)^-1$

To find a max/min we find the first derivative and find the values for which the derivative is zero.

$y = {(sin x)}^{- 1}$ $y=(sinx)^-1$
$:.y'=(-1)(sinx)^-2(cosx)$ (chain rule)
$:.y'=-cosx/sin^2x$

At max/min, $y'=0=>-cosx/sin^2x=0$
$:.cosx=0$
$:.x=-pi/2,pi/2, ...$

When $x=pi/2=>y=1/sin(pi/2)=1$
When $x=-pi/2=>y=1/sin(-pi/2)=-1$

So there are turning points at $(-pi/2,-1)$ and $(pi/2,1)$

If we look at the graph of $y=cscx$ the we observe that $(-pi/2,-1)$ is a relative maximum and $(pi/2,1)$ is a relative minimum.

graph{csc x [-4, 4, -5, 5]}