Question

consider the function $f(x)=\cos(x)-\frac{\sqrt{3}}{2}x$ this function has two critical numbers $a < b$ in $0,2\pi$. give the following:
$a=\square$
$b=\square$
$f(a)=\square$
$f(b)=\square$
thus $f(x)$ has a local $? <$ at $a$ and a local $? <$ at $b$.

Explanation:

Step1: Find the first - derivative

The function is \(f(x)=\cos(x)-\frac{\sqrt{3}}{2}x\).
The first - derivative \(f^{\prime}(x)=-\sin(x)-\frac{\sqrt{3}}{2}\).
Set \(f^{\prime}(x) = 0\), then \(-\sin(x)-\frac{\sqrt{3}}{2}=0\), so \(\sin(x)=-\frac{\sqrt{3}}{2}\).
Since \(x\in[0,2\pi]\), \(x = \frac{4\pi}{3}\) or \(x=\frac{5\pi}{3}\). So \(A=\frac{4\pi}{3}\), \(B = \frac{5\pi}{3}\).

Step2: Find the second - derivative

The second - derivative \(f^{\prime\prime}(x)=-\cos(x)\).
When \(x = A=\frac{4\pi}{3}\), \(f^{\prime\prime}(\frac{4\pi}{3})=-\cos(\frac{4\pi}{3})=\frac{1}{2}\).
When \(x = B=\frac{5\pi}{3}\), \(f^{\prime\prime}(\frac{5\pi}{3})=-\cos(\frac{5\pi}{3})=-\frac{1}{2}\).

Step3: Determine local extrema

By the second - derivative test:
If \(f^{\prime\prime}(c)>0\), then \(f(x)\) has a local minimum at \(x = c\). If \(f^{\prime\prime}(c)<0\), then \(f(x)\) has a local maximum at \(x = c\).
Since \(f^{\prime\prime}(A)=\frac{1}{2}>0\), \(f(x)\) has a local minimum at \(A\).
Since \(f^{\prime\prime}(B)=-\frac{1}{2}<0\), \(f(x)\) has a local maximum at \(B\).

Answer:

\(A=\frac{4\pi}{3}\), \(B=\frac{5\pi}{3}\), \(f^{\prime\prime}(A)=\frac{1}{2}\), \(f^{\prime\prime}(B)=-\frac{1}{2}\), local minimum at \(A\), local maximum at \(B\).