How Do You Find Relative Extrema of the Function h(x) = cos(π/x)?

[k3k3]

Homework Statement

Let h:(0,1)→ℝ be defined by h(x)=cos($\pi\\$/x). Determine the set of all x such that x has a relative extreme value at x.

Homework Equations

The Attempt at a Solution

It is common knowledge that the min and max are -1 and 1. Using intuition, it is clear by guess and check to arrive at x=1/(2n+1) gives a minimum and x=1/2n gives a maximum.

Using the first derivative test, I arrive at 1=0 which is not true. I am unsure how to interpret this answer.

Trying to solve $d^{2}$(cos($\pi\\$/x))/d$x^{2}$=0 is proving to be a real challenge.

I am only concerned about this since my professor wants me to show my work, so I do not think saying I guessed and checked will be acceptable.

 

[Mark44]

Homework Statement

Let h:(0,1)→ℝ be defined by h(x)=cos($\pi\\$/x). Determine the set of all x such that x has a relative extreme value at x.

Homework Equations

The Attempt at a Solution

It is common knowledge that the min and max are -1 and 1. Using intuition, it is clear by guess and check to arrive at x=1/(2n+1) gives a minimum and x=1/2n gives a maximum.

Using the first derivative test, I arrive at 1=0 which is not true. I am unsure how to interpret this answer.

Trying to solve $d^{2}$(cos($\pi\\$/x))/d$x^{2}$=0 is proving to be a real challenge.

I am only concerned about this since my professor wants me to show my work, so I do not think saying I guessed and checked will be acceptable.

Instead of guessing at the answer, find the values of x for which h'(x) = 0.

Show us what you did to arrive at 1 = 0.

 

[k3k3]

Instead of guessing at the answer, find the values of x for which h'(x) = 0.

Show us what you did to arrive at 1 = 0.

Using the chain rule, h'(x)=$\pi\\$sin($\pi\\$/x)*1/x$^{2}$

Set it equal to 0

$\pi\\$sin($\pi\\$/x)*1/x$^{2}$=0

Multiply both sides by x$^{2}$/$\pi\\$ yields sin($\pi\\$/x)=0

Take the arcsin of both sides yields

$\pi\\$/x=0

Divide both sides by pi,

1/x=0

Multiply both sides by x

1=0

Or am I forgetting that there are more possible values that arcsin can be 0?

 

[Ray Vickson]

Using the chain rule, h'(x)=$\pi\\$sin($\pi\\$/x)*1/x$^{2}$

Set it equal to 0

$\pi\\$sin($\pi\\$/x)*1/x$^{2}$=0

Multiply both sides by x$^{2}$/$\pi\\$ yields sin($\pi\\$/x)=0

Take the arcsin of both sides yields

$\pi\\$/x=0

Divide both sides by pi,

1/x=0

Multiply both sides by x

1=0

Or am I forgetting that there are more possible values that arcsin can be 0?

You should not be looking at the arcsin; you should be looking for solutions of the equation sin(w) = 0. Can you see why the arcsin misses all but one solution?

RGV