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SOLUTION 14: We are given the equation
x3=2x2+3x−3 ⟶ x3−2x2−3x+3=0
Let function
f(x)=x3−2x2−3x+3 and choose m=0
This function is continuous for all values of x since it is a polynomial. We know from algebra that a cubic polynomial has at most three distinct real roots.
To establish three distinct and appropriate intervals, consider the graph of this function.
INTERVAL ONE: Note that f(0)=(0)3−2(0)2−3(0)+3=3<0 and f(−1)=(−2)3−2(−2)2−3(−2)+3=−7<0
so that f(−2)=−7<m<3=f(0)
i.e., m=0 is between f(−2) and f(0). Now choose the interval to be [−2,0].
The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number c in the interval [−π,0] which satisfies
f(c)=m
i.e.,
c3−2c2−3c+3=0
and the equation is solvable on this interval.
INTERVAL TWO: Note that f(0)=(0)3−2(0)2−3(0)+3=3<0 and f(1)=(1)3−2(1)2−3(1)+3=−1<0
so that f(1)=−1<m<3=f(0)
i.e., m=0 is between f(0) and f(1). Now choose the interval to be [0,1].
The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number c in the interval [0,1] which satisfies
f(c)=m
i.e.,
c3−2c2−3c+3=0
and the equation is solvable on this interval.
INTERVAL THREE: Note that f(3)=(3)3−2(3)2−3(3)+3=3>0 and f(1)=(1)3−2(1)2−3(1)+3=−1<0
so that f(1)=−1<m<3=f(3)
i.e., m=0 is between f(1) and f(3). Now choose the interval to be [1,3].
The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number c in the interval [1,3] which satisfies
f(c)=m
i.e.,
c3−2c2−3c+3=0
and the equation is solvable on this interval. This completes the solution.
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