Question

(a) find the interval on which f is increasing. find the interval on which f is decreasing.
for ( f(x)=4 sin (x)+4 cos (x) ), we have
( f^{prime}(x)=4 cos (x)-4 sin (x) )
if this equals 0, then we have ( cos (x)=sin (x) ), which becomes ( \tan (x)=1 ). hence, in the interval ( 0 leq x leq 2 pi, f^{prime}(x)=0 ) when ( x=\frac{pi}{4} ) or ( x=\frac{5 pi}{4} )
step 2
if ( f^{prime}(x) ) is negative, then ( f(x) ) is decreasing. if ( f^{prime}(x) ) is positive, then ( f(x) ) is increasing
step 3
if ( 0<x<\frac{pi}{4} ) then ( f^{prime}(x) ) is -select-, and ( f(x) ) is -select-

Explanation:

Step1: Analyze the sign of \(f^{\prime}(x)\) in the interval \(0 < x<\frac{\pi}{4}\)

Take a test - value \(x = \frac{\pi}{6}\) in the interval \(0 < x<\frac{\pi}{4}\).
\(f^{\prime}(x)=4\cos(x)-4\sin(x)\), substitute \(x = \frac{\pi}{6}\)
\(f^{\prime}(\frac{\pi}{6})=4\cos(\frac{\pi}{6})-4\sin(\frac{\pi}{6})\)
Since \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\) and \(\sin(\frac{\pi}{6})=\frac{1}{2}\)
\(f^{\prime}(\frac{\pi}{6})=4\times\frac{\sqrt{3}}{2}-4\times\frac{1}{2}=2\sqrt{3}-2\approx2\times1.732 - 2=3.464 - 2=1.464>0\)

Step2: Determine the behavior of \(f(x)\)

Since \(f^{\prime}(x)>0\) for \(0 < x<\frac{\pi}{4}\) (from Step 1), by the first - derivative test, if \(f^{\prime}(x)>0\) on an interval, then \(f(x)\) is increasing on that interval.

Answer:

If \(0 < x<\frac{\pi}{4}\), then \(f^{\prime}(x)\) is positive, and \(f(x)\) is increasing.