Question

analyze the following functions. which function has relative extrema? (1 point)
$f(x)=4x - 6.7$
$f(x)=sin(x - 3)+5$
$f(x)=1.4\tan(x + 2)$
$f(x)=sqrt{50x}$

Explanation:

Step1: Analyze \( f(x)=4x - 6.7 \)

The function \( f(x)=4x - 6.7 \) is a linear function. Its derivative \( f^{\prime}(x)=4\). Since \( f^{\prime}(x)
eq0 \) for all \( x\), it has no relative extrema.

Step2: Analyze \( f(x)=\sin(x - 3)+5 \)

The derivative of \( f(x)=\sin(x - 3)+5 \) is \( f^{\prime}(x)=\cos(x - 3)\). Set \( f^{\prime}(x) = 0\), then \(\cos(x - 3)=0\). Solving \(x-3=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\), \(x = 3+\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\). The second - derivative \( f^{\prime\prime}(x)=-\sin(x - 3)\). When \(x = 3+\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}\), \(f^{\prime\prime}(x)=- 1<0\) (relative maximum); when \(x = 3+\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z}\), \(f^{\prime\prime}(x)=1>0\) (relative minimum).

Step3: Analyze \( f(x)=1.4\tan(x + 2) \)

The derivative \( f^{\prime}(x)=1.4\sec^{2}(x + 2)\). Since \(\sec^{2}(x + 2)=\frac{1}{\cos^{2}(x + 2)}\geq1\) for all \(x\) where \(\cos(x + 2)
eq0\), \(f^{\prime}(x)=1.4\sec^{2}(x + 2)>0\) for all \(x\) in the domain of \(f(x)\). So it has no relative extrema.

Step4: Analyze \( f(x)=\sqrt{50x}\)

The function \( f(x)=\sqrt{50x}=5\sqrt{2}\sqrt{x}\), its domain is \(x\geq0\). The derivative \( f^{\prime}(x)=\frac{5\sqrt{2}}{2\sqrt{x}}\). For \(x > 0\), \(f^{\prime}(x)>0\), and at \(x = 0\), the derivative from the right is \(+\infty\). So it has no relative extrema.

Answer:

\( f(x)=\sin(x - 3)+5 \)