Question

On what intervals within [0, 2π) is f increasing? f(x) = sin (x + π/6)

Answer 1

find f'(x) > 0

Answer 2

Perhaps between the min and the max?

Answer 3

I was thinking somewhere between [0, 5pi/12] U [11pi/12, 11pi/6] but its not working.. So i'm doing something wrong, but I don't know what.

Answer 4

hmmm do you know the graph of sin(x) ?

Answer 5

for f(x) to be increasing,f'(x)>0
\[f \prime \left( x \right)=\cos \left( x+\frac{ \pi }{ 6 } \right)>0\]
\[\cos \left( x+\frac{ \pi }{ 6 } \right)\ge 0,~or~x+\frac{ \pi }{6 }\in \left( 0,\frac{ \pi }{ 2 } \right)\cup \left( \frac{ 3\pi }{2 },2\pi \ \right)\]
\[x \in \left( \frac{ -\pi }{6 },\frac{ \pi }{2 }-\frac{ \pi }{ 6 } \right)\cup \left( \frac{ 3\pi }{ 2 }-\frac{ \pi }{6 },\left( 2\pi-\frac{ \pi }{6 } \right) \right)\]
\[x \in \left( \frac{ -\pi }{ 6 },\frac{ \pi }{ 3 } \right)\cup \left(\frac{ 4\pi }{ 3 },\frac{ 11\pi }{ 6 } \right)\]

Answer 6

\(\bf \textit{parent function}\implies f(x)=sin(x)\\ \quad \\

\begin{array}{llll}
f(x)=sin&\left(x{\color{red}{ +\frac{\pi}{6}}}\right)\\
&\qquad \uparrow \\
&\textit{horizontal shift to the left}
\end{array}\)

Answer 7

Yea i know the graph of sin(x) but since this is like a phase shift problem doesnt that move everything over to the left. Because its a positive "c" value. like everything shifts by -pi/6 to the left. and since its between [0,2pi] would i still include the point [0, 5pi/12] as one of the increasing intervals? I get what it is supposed to do, but i somewhere im doing something wrong as far as actually figuring out what the intervals are supposed to be.

Answer 8

I really appreciate everyone's feedback though. It has all been very helpful!

Answer 9

May I assume you want to continue with the solution of this problem?

Answer 10

Yes haha, the reply from surjithayer was very close, but it has to be between [0,2pi) and it won't work if it isnt. I tried using 0 instead of what he had, but it still wasnt working. I honestly don't know what i am doing wrong at this point

Answer 11

Sirji has determined that the derivative of f(x)=sin(x + pi/6) is simply f '(x)=cos(x+pi/6); I agree. As he indicated, we need to determine on which intervals cos(x+pi/6) > 0, which is/are where f(x) is increasing.

Use the sum formula for cos (a + b) = cos a cos b - sin a sin b
to obtain cos (x+pi/6) = cos x cos pi/6 - sin x sin pi/6,
or \[\cos (x+\frac{ \pi }{ 6 })=(\cos x)(\frac{ \sqrt{3} }{ 2 }) - (\sin x)(\frac{ 1 }{ 2 }) > 0\]
Solving this inequality will lead to identification of the intervals on which the original function is increasing.

Answer 12

when we find the interval as mr.mathmale suggested ,we can not divide by cos x or sin x ,because we are not clear they are positive or negative and negative sign changes the inequality.
In my opinion
\[x \in \left( 0,\frac{ \pi }{ 3 } \right)\cup \left( \frac{ 4\pi }{ 3 } ,\frac{ 11\pi }{ 6 }\right)\]