Question

a. prove that f(x)=x - ln x is increasing for x > 1. b. using part (a), show that ln x < x for x > 1. what is/are the critical point(s)? x = 1 (use a comma to separate answers as needed.) what is the next step? a. substitute the critical point(s) in f to determine the sign of f on that interval. b. substitute values on either side of the critical point(s) in f to determine the sign of f on that interval. c. substitute values on either side of the critical point(s) in f to determine the sign of f on that interval. d. substitute the critical point(s) in f to determine the sign of f on that interval. therefore, f(x)=x - ln x is increasing for x > 1 because the sign of is on the interval (type your answer using interval notation.)

Explanation:

Step1: Find the derivative of \(f(x)\)

The derivative of \(y = f(x)=x-\ln x\) is \(f'(x)=\frac{d}{dx}(x)-\frac{d}{dx}(\ln x)=1 - \frac{1}{x}=\frac{x - 1}{x}\), for \(x>0\).

Step2: Find the critical points

Set \(f'(x) = 0\), so \(\frac{x - 1}{x}=0\). Since \(x>0\), then \(x - 1=0\), and the critical - point is \(x = 1\).

Step3: Determine the sign of \(f'(x)\) for \(x>1\)

For \(x>1\), if we take a value \(x=a>1\), then \(f'(a)=\frac{a - 1}{a}\). Since \(a>1\), \(a - 1>0\) and \(a>0\), so \(f'(a)>0\). When the derivative \(f'(x)>0\) on an interval, the function \(y = f(x)\) is increasing on that interval. So \(f(x)=x-\ln x\) is increasing for \(x > 1\).

Step4: Prove \(\ln x1\)

Since \(f(x)=x-\ln x\) is increasing for \(x > 1\), then for \(x>1\), \(f(x)>f(1)\). Calculate \(f(1)=1-\ln1=1\). So \(f(x)=x-\ln x>1>0\) for \(x > 1\). Rearranging the inequality \(x-\ln x>0\) gives \(\ln x1\).

Answer:

a. We found \(f'(x)=\frac{x - 1}{x}\), the critical - point is \(x = 1\), and for \(x>1\), \(f'(x)>0\), so \(f(x)\) is increasing for \(x>1\).
b. Since \(f(x)=x-\ln x\) is increasing for \(x > 1\) and \(f(1) = 1\), then \(x-\ln x>0\) for \(x>1\), which implies \(\ln x1\).