f(x)=x3−3x+5f(−2)=(−2)3−3×(−2)+5=−8+6+5=3>0f(−3)=(−3)3−3×(−3)+5=−27+9+5=−13<0f(x)= x^3-3x+5 \\f(-2)=(-2)^3-3\times(-2)+5=-8+6+5=3>0 \\f(-3)=(-3)^3-3\times(-3)+5=-27+9+5=-13<0f(x)=x3−3x+5f(−2)=(−2)3−3×(−2)+5=−8+6+5=3>0f(−3)=(−3)3−3×(−3)+5=−27+9+5=−13<0
As f(−2)>0,f(−3)<0f(-2)>0 , f(-3)<0f(−2)>0,f(−3)<0
Then from IVT, there will be a root between [−2,−3].[-2,-3].[−2,−3].
This is only interval of length 1 for which , we will get the roots or solution of f(x).f(x).f(x).
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