SOLUTION: Given the function {{{ f(x) = x^3-3x+2 }}} on the domain {{{ a<=x<=a+2}}} , find the value of a at which two ends of the domain give the same value , equal to the maximum value of

Question 1036690: Given the function on the domain , find the value of a at which two ends of the domain give the same value , equal to the maximum value of the interval.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The graph of the equation f(x) = x^3 -3x + 2 is
:

:
note that x^3 -3x +2 = (x-1)^2 * (x+2)
:
we are looking for the value of a that satisfies
:
a^3 -3a + 2 = (a+2-1)^2 * (a+2+2)
:
a^3 -3a +2 = a^3 +6a^2 +9a +4
:
6a^2 +12a +2 = 0
:
a^2 +2a + 1/3 = 0
:
a^2 +2a + 1 = (-1/3) + 1
:
(a+1)^2 = 2/3
:
a = -1 + square root(2/3) = -0.18
a = -1 - square root(2/3) = -1.82
:
check these values
:
f(-0.18) = 2.53
f(1.82) = 2.53
:
*************************
interval is [-0.18, 1.82]
*************************
:
note the interval [-1.82, 0.18] has the local max 4 at a=-1, therefore we exclude this interval
:

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