Question

Algebraically find the absolute maximum and minimum of f(x)=x-3ln(x) over [1,4]

Answer 1

Derive and set equal to zero:

0 = 1 - 3/x

x = 3

Checking whether it's a min or max by calculating second derivative:

f''(x) = 3/x^2 > 0 therefore it's a minimum.

Subbing in the equation:

f(3) = 3 - 3ln(3) = -0.295

Now checking the endpoints:

f(1) = 1- 3ln1 = 1

f(4) = 4 - 3ln4 = -.158

So abs min: (3, -0.295)
max: (1,1)