What are the absolute extrema of #f(x)=(x^3-7x^2+12x-6)/(x-1)in[1,4]#?

1 Answer
Apr 15, 2018

Answer:

There's no global maxima.

The global minima is -3 and occurs at x = 3.

Explanation:

#f(x) = (x^3 - 7x^2 + 12x - 6)/(x - 1)#

#f(x) = ((x - 1)(x^2 - 6x + 6)) /(x - 1)#

#f(x) = x^2 - 6x + 6, #where #x≠ 1#

#f'(x) = 2x - 6#

The absolute extrema occurs on an endpoint or at the critical number.

Endpoints: #1 & 4: #

#x = 1#
# f(1) : "undefined" #
#lim_(x→1) f(x) = 1#

#x = 4#
# f(4) = -2 #

Critical point(s):
#f'(x) = 2x - 6#
# f'(x) = 0#
#2x - 6 = 0 , x = 3#

At # x = 3 #
# f(3) = -3 #

There's no global maxima.

There's no global minima is -3 and occurs at x = 3.