Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = x − ln 8x, [1/2, 2]

The given function is 
f(x) =  x - ln(8x), on the interval [1/2, 2].

The derivative of f is
f'(x) = 1 - 1/x
The second derivative is 
f''(x) = 1/x²

A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0  => 1/x = 1  => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079

The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2)  = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.

 
Minimum value  = 1 - ln(8)
Maximum value = 2 - ln(16)

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