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.

Answer:

Minimum value = 1 - ln (8)

Maximum value = 2 - ln (16)