- #1
TitoSmooth
- 158
- 6
I was stuck for an hour trying to do this calculus 1 problem. Think I figured it out but it's a even problm.
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
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