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

1 Answer
Nov 5, 2015

Answer:

#(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]}