Question

over which interval is the graph of ( y = cos(x) ) strictly increasing?
( 0 < x < \frac{pi}{2} )
( 0 < x < pi )
( \frac{pi}{2} < x < \frac{3pi}{2} )
( pi < x < 2pi )

Explanation:

Step1: Recall the derivative of \(y = \cos(x)\)

The derivative of \(y=\cos(x)\) is \(y'=-\sin(x)\). A function \(y = f(x)\) is strictly increasing when \(y'>0\). So we need to find when \(-\sin(x)>0\), which is equivalent to \(\sin(x)<0\).

Step2: Analyze the sign of \(\sin(x)\) in different intervals

• For \(0 < x<\frac{\pi}{2}\), \(\sin(x)>0\) (since in the first - quadrant, sine function is positive).
• For \(0 < x<\pi\), \(\sin(x)>0\) when \(0 < x<\pi\) (sine is positive in the first and second quadrants).
• For \(\frac{\pi}{2}
• For \(\pi

Answer:

\(\pi < x<2\pi\)