we have $f(x)=4cos(x)-4sin(x)$, so $f(x)=square$, which equals 0 when $\tan(x)=square$. hence, in the…

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 $x=\frac{7pi}{4}$.

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 $x=\frac{7pi}{4}$.

Answer

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\neq0)). 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)\neq0)). 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)