Question

determine the absolute extreme values of the function f(x)=sinx-cosx+6 on the interval 0<=x<=2pi.
can you please explain how to answer this question?

Answer 1

differentiate it and find x in \(0<=x<=2\pi\)
find 2nd order derivative at these points for maxima and then find max. value of f(x)

Answer 2

I found the derivative which is f'(x)=cosx+sinx. how would I find x?

Answer 3

put f'(x)=0

Answer 4

f'(x)=0
0=cosx+sinx
=x(cos+sin)
is that correct? im not sure what to do afterwards.

Answer 5

no

Answer 6

\[\sin x+\cos x=0,\sin x=-\cos x\]
\[\frac{ \sin x }{ \cos x }=-1\]
\[\tan x=-1=-\tan \frac{ \pi }{ 4 }=\tan \left( \pi-\frac{ \pi }{ 4 } \right),\tan \left( 2 \pi-\frac{ \pi }{ 4 } \right)\]
\[x=\frac{ 3 \pi }{ 4 },\frac{ 7 \pi }{ 4 }\]

Answer 7

oh ok thanks! I have another question, where did the third step come from? The inverse of tanx= -1 which is -45 degrees which is pi/4 in radians but where did tan(pi-pi/4), tan(2pi-pi/4) come from?

Answer 8

\[\tan(\pi- \theta)=-\tan \theta,\tan(2 \pi-\theta)=-\tan \theta\]
\[\tan(\pi-\theta)=\frac{ \tan \pi-\tan \theta }{ 1+\tan \pi \tan \theta }=\frac{ 0-\tan \theta }{ 1+0\times \tan \theta }=-\tan \theta\]
similarly \(tan(2 \pi-\theta\))

Answer 9

Oh ok also the unit circle can be used right?

Answer 10

correct.