Therefore absolute max exists at either interval endpoint.
f(1) = 2.5 and f(6) = 10/3.
Hence max value of f(x) is 10/3, which can be found at x = 6.
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Alternatively, knowing that for a,b > 0
===> , with equality happening only for x = 2.
Hence there is minimum of 2 at x = 2. Absolute max then exists at either interval endpoint of [1,6],
and we get the same conclusion as in previous calculus approach.