Question 980385
f(x) = ln(x)/x^2 on [1,3]
Since the denominator is always positive, and the minimum value of ln(x) on this interval occurs at x=1: ln(1) = 0
The maximum value is obtained by taking the derivative and setting=0:
df/dx = 0 = 1/x^3 - 2*ln(x)/x^3 -> 2*ln(x) = 1 -> ln(x) = 1/2
exp(ln(x)) = x = exp(1/2)
f(exp(1/2) = ln(exp(1/2)/(exp(1/2))^2 = (1/2)/(exp(1/2))^2 (~0.184)
So minimum value = 0 and maximum value = (1/2)/(exp(1/2))^2 (~0.184)