Question

consider the function f(x)=cos x-sin x on the domain [-pi,pi]. Use the first derivative test to find the open intervals on which the function is decreasing.

Answer 1

I got decreasing on (-pi/4,3pi/4) but not sure if this is correct?

Answer 2

f(x) = cos x - sinx
f'(x) = -sinx - cosx = -(sinx + cosx)
f'(x) = 0 requires -(sinx + cosx) = 0
I would use the transformation formula to make sinx + cosx = \[\sqrt{2}\sin (x+\frac{ \pi }{ 4 })\]
When is \[\sqrt{2}\sin (x+\frac{ \pi }{ 4 })\] negative? this would be when
\[-\pi<x+\frac{ \pi }{ 4 } < 0 \] or when \[\pi<x+\frac{ \pi }{ 4 } < 2\pi \]
But now we have our domain to consider:
\[Dom(f) = -\pi<x<\pi \]
So our new domain is \[-\pi + \frac{ \pi }{ 4 }< x + \frac{ \pi }{ 4 }< \pi + \frac{ \pi }{ 4 }\]
whichis \[\frac{ -3\pi }{ 4 } <x+ \frac{ \pi }{ 4 }< \frac{ 5\pi }{ 4 }\]
So returning to our original
When is \[\sqrt{2}\sin (x+\frac{ \pi }{ 4 })\] negative? this would be when
\[-\pi<x+\frac{ \pi }{ 4 } < 0 \] or when \[\pi<x+\frac{ \pi }{ 4 } < 2\pi \]
i.e when \[\frac{ -5\pi }{ 4 } < x < \frac{ -\pi }{ 4 } \]
or \[\frac{3\pi }{ 4 } < x < \frac{ 7\pi }{ 4 } \]
Given the domain restrictions we finally have\[\frac{ -3\pi }{ 4 } < x < \frac{ -\pi }{ 4 } \cup \frac{ 3\pi }{ 4 } < x< \frac{ 5\pi }{ 4 } \]

Answer 3

sorry I misread on the domain restrictions. The domain is still \[-\pi<x<\pi \] so the answer would be \[-\pi<x<\frac{ -\pi }{ 4 } \cup \frac{ 3\pi }{ 4 } < x< \pi\]

Answer 4

A plot and solution using Mathematica is attached.