Question

we have $f(x)=4cos(x)-4sin(x)$, so $f(x)=square$, which equals 0 when $\tan(x)=square$. hence, in the interval $0leq xleq2pi$, $f(x)=0$ when $x=\frac{3pi}{4}$ and $x=\frac{7pi}{4}$.

Explanation:

Step1: Find the second - derivative

We know that if \(y = \cos(x)\), then \(y^\prime=-\sin(x)\) and if \(y=\sin(x)\), then \(y^\prime = \cos(x)\).
Given \(f^\prime(x)=4\cos(x)-4\sin(x)\).
Using the sum - rule of differentiation \((u - v)^\prime=u^\prime - v^\prime\) where \(u = 4\cos(x)\) and \(v = 4\sin(x)\).
The derivative of \(u = 4\cos(x)\) with respect to \(x\) is \(u^\prime=-4\sin(x)\) (since \((a\cos(x))^\prime=-a\sin(x)\) for \(a = 4\)), and the derivative of \(v = 4\sin(x)\) with respect to \(x\) is \(v^\prime = 4\cos(x)\) (since \((a\sin(x))^\prime=a\cos(x)\) for \(a = 4\)).
So \(f^{\prime\prime}(x)=\frac{d}{dx}(4\cos(x)-4\sin(x))=-4\sin(x)-4\cos(x)\).

Step2: Solve \(f^{\prime\prime}(x)=0\)

Set \(f^{\prime\prime}(x)=-4\sin(x)-4\cos(x) = 0\).
Divide both sides of the equation \(-4\sin(x)-4\cos(x)=0\) by \(- 4\) (since \(-4
eq0\)). We get \(\sin(x)+\cos(x)=0\).
Rearrange the equation \(\sin(x)+\cos(x)=0\) to \(\sin(x)=-\cos(x)\).
Then \(\tan(x)=\frac{\sin(x)}{\cos(x)}=- 1\) (assuming \(\cos(x)
eq0\)).
We know that \(\tan(x)=-1\) when \(x = n\pi-\frac{\pi}{4},n\in\mathbb{Z}\).
For the interval \(0\leq x\leq2\pi\), when \(n = 1\), \(x=\frac{3\pi}{4}\) and when \(n = 2\), \(x=\frac{7\pi}{4}\).

Answer:

\(f^{\prime\prime}(x)=-4\sin(x)-4\cos(x)\), \(\tan(x)=-1\)