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…
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?\na f is decreasing for x>1, and the graph of f is concave down for x>e.\nb f is decreasing for x>1, and the graph of f is concave up for x>e.\nc f is increasing for x>1, and the graph of f is concave down for x>e.\nd f is increasing for x>1, and the graph of f is concave up for x>e.\ne f is increasing for 0<x<e, and the graph of f is concave down for 0<x<e^{3/2}
Answer
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)