- #1

- 158

- 6

Find the absolute maximum and absolute minimum values of f on the given interval.

f(t)=t+cot (t/2), [pie/4,7pie/4]

f'=1-(1/2) csc^2 (t/2)

So 1=1/2*csc^2 (t/2)

2=csc^2 (t/2)

For some reason I didn't know what to to do with cc squared so I applied an identity.

+/_ 1=cot (t/2)

Took arc cot on both side.

Arccot (+-1) =t/2

The arc cot gives me pie/4 and 3pie/4

So now I multiply both by 2.

So I get pie/2=t or 3pie/2=t.

Then I just plug the values in f and solve.

To see which ones are global max n min including the endpoints.

Also my question was.

Are we allowed to take the inverse cot on both sides of

Arccot (1)=arccot (cot^2 (t/2)) ?

If so how would we work with it?

I was thinking arccot1=1 and on the right side. Arccot will cancel one cot so I'm left with cot(t/2)

Now 1=cot (t/2)

Then I use pie/4 +pie (k)=t/2 and solve for solutions in the restricted interval? Is this correct