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)=)
if this equals 0, then we have ( cos (x)=), which becomes ( \tan (x)=). hence, in the interval ( 0 leq x leq 2 pi, f^{prime}(x)=0 ) when ( x=\frac{pi}{4} ) or
( x=)

Explanation:

Step1: Find the derivative of \(f(x)\)

Using the derivative rules \((\sin x)'=\cos x\) and \((\cos x)' =-\sin x\).
For \(f(x)=4\sin(x)+4\cos(x)\), by the sum - rule \((u + v)'=u'+v'\) where \(u = 4\sin(x)\) and \(v = 4\cos(x)\)
\(f'(x)=4\cos(x)-4\sin(x)\)

Step2: Set \(f'(x) = 0\) and simplify

If \(4\cos(x)-4\sin(x)=0\), then divide both sides by \(4\) (since \(4
eq0\))
\(\cos(x)-\sin(x)=0\), which gives \(\cos(x)=\sin(x)\)
Divide both sides by \(\cos(x)\) (assuming \(\cos(x)
eq0\)). We know that \(\tan(x)=\frac{\sin(x)}{\cos(x)}\), so \(\tan(x) = 1\)
The general solution of \(\tan(x)=1\) is \(x = n\pi+\frac{\pi}{4},n\in\mathbb{Z}\). In the interval \(0\leq x\leq2\pi\), when \(n = 0\), \(x=\frac{\pi}{4}\) and when \(n = 1\), \(x=\frac{5\pi}{4}\)

Answer:

\(f'(x)=4\cos(x)-4\sin(x)\); \(\cos(x)=\sin(x)\); \(\tan(x)=1\); \(x = \frac{5\pi}{4}\)