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

Algebra ->  Points-lines-and-rays -> 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       Log On


   

Question 1036690: Given the function +f%28x%29+=+x%5E3-3x%2B2+ on the domain +a%3C=x%3C=a%2B2 , 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) About Me  (Show Source):
You can put this solution on YOUR website!
The graph of the equation f(x) = x^3 -3x + 2 is
:
+graph%28+300%2C+200%2C+-2.5%2C+2.5%2C+-5%2C+10%2C+x%5E3+-3x+%2B+2%29+
:
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
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check these values
:
f(-0.18) = 2.53
f(1.82) = 2.53
:
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interval is [-0.18, 1.82]
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note the interval [-1.82, 0.18] has the local max 4 at a=-1, therefore we exclude this interval
: