Question

for x>0, f is a function such that f(x)=\frac{\ln x}{x} and f(x)=\frac{1 - \ln x}{x^{2}}. which of the following is true?
a f is decreasing for x>1, and the graph of f is concave down for x>e.
b f is decreasing for x>1, and the graph of f is concave up for x>e.
c f is increasing for x>1, and the graph of f is concave down for x>e.
d f is increasing for x>1, and the graph of f is concave up for x>e.
e f is increasing for 0<x<e, and the graph of f is concave down for 0<x<e^{3/2}

Explanation:

Step1: Analyze the first - derivative for increasing/decreasing

The first - derivative of the function is \(f^{\prime}(x)=\frac{\ln x}{x}\).
For \(x > 1\), when \(x>1\), \(\ln x>0\) (since \(y = \ln x\) is an increasing function with \(\ln1 = 0\)) and \(x>0\). So \(f^{\prime}(x)=\frac{\ln x}{x}>0\) for \(x > 1\) (because the numerator \(\ln x>0\) and the denominator \(x>0\)). A function \(y = f(x)\) is increasing when \(f^{\prime}(x)>0\) and decreasing when \(f^{\prime}(x)<0\).

Step2: Analyze the second - derivative for concavity

The second - derivative of the function is \(f^{\prime\prime}(x)=\frac{1-\ln x}{x^{2}}\).
For \(x>e\), \(\ln x>\ln e = 1\). Then \(1-\ln x<0\) and \(x^{2}>0\) (since \(x>0\)). So \(f^{\prime\prime}(x)=\frac{1 - \ln x}{x^{2}}<0\) for \(x>e\). A function \(y = f(x)\) is concave down when \(f^{\prime\prime}(x)<0\) and concave up when \(f^{\prime\prime}(x)>0\).

Answer:

C. \(f\) is increasing for \(x > 1\), and the graph of \(f\) is concave down for \(x > e\)