What are the extrema of f(x)=x^2 - 6x + 11 on x in[1,6]?

1 Answer
Nov 5, 2015

(3,2) is a minimum.
(1,6) and (6,11) are maxima.

Explanation:

Relative extrema occur when f'(x)=0.
That is, when 2x-6=0.
ie when x=3.

To check if x=3 is a relative minimum or maximum, we observe that f''(3)>0 and so => x=3 is a relative minimum,
that is, (3, f(3))=(3,2) is a relative minimum and also an absolute minimum since it is a quadratic function.

Since f(1)=6 and f(6)=11, it implies that (1,6) and (6,11) are absolute maxima on the interval [1,6].

graph{x^2-6x+11 [-3.58, 21.73, -0.37, 12.29]}