Consider the function $f(x)=\frac{\ln (x)}{x^{2}}$. For this function there are two important numbers $A
Solution
Step 1 ：Consider the function \(f(x)=\frac{\ln (x)}{x^{2}}\).
Step 2 ：Find the derivative of the function \(f(x)\), \(f'(x) = -2\ln(x)/x^{3} + x^{-3}\).
Step 3 ：Determine the critical points of \(f(x)\) by setting \(f'(x)\) equal to zero or undefined. The critical point is \(x = e^{1/2}\).
Step 4 ：Consider the domain of the original function, which is \((0, \infty)\). Therefore, the two important numbers are \(A=0\) and \(B=e^{1/2}\).
Step 5 ：Determine the intervals where the derivative is positive or negative. The derivative is positive in the interval \((0, e^{1/2})\) and negative in the interval \((e^{1/2}, \infty)\).
Step 6 ：Conclude that the function \(f(x)\) is increasing in the first interval and decreasing in the second interval. Therefore, \(f(x)\) has a local maximum at \(x = e^{1/2}\).
Step 7 ：\(\boxed{A=0, B=e^{1/2}, f'(x)\) is positive in \((A, B)\) and negative in \((B, \infty)\). Thus, \(f(x)\) has a local MAX at \(B\).}
From Solvely APP
Solution
Step 1 ：Consider the function \(f(x)=\frac{\ln (x)}{x^{2}}\).
Step 2 ：Find the derivative of the function \(f(x)\), \(f'(x) = -2\ln(x)/x^{3} + x^{-3}\).
Step 3 ：Determine the critical points of \(f(x)\) by setting \(f'(x)\) equal to zero or undefined. The critical point is \(x = e^{1/2}\).
Step 4 ：Consider the domain of the original function, which is \((0, \infty)\). Therefore, the two important numbers are \(A=0\) and \(B=e^{1/2}\).
Step 5 ：Determine the intervals where the derivative is positive or negative. The derivative is positive in the interval \((0, e^{1/2})\) and negative in the interval \((e^{1/2}, \infty)\).
Step 6 ：Conclude that the function \(f(x)\) is increasing in the first interval and decreasing in the second interval. Therefore, \(f(x)\) has a local maximum at \(x = e^{1/2}\).
Step 7 ：\(\boxed{A=0, B=e^{1/2}, f'(x)\) is positive in \((A, B)\) and negative in \((B, \infty)\). Thus, \(f(x)\) has a local MAX at \(B\).}
