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Intermediate value theorem — root of equation

Show that the equation x³ − 3x + 1 = 0 has at least one real root in the interval (1, 2) using Bolzano's theorem.
Given data
f(x) = x³ − 3x + 1Interval (1, 2)
Review the theory: Continuità
Steps
0 / 3
  1. Calculate f(1).
  2. Calculate f(2).
  3. Do f(1) and f(2) have opposite signs? What does Bolzano say?
Full worked solution
  1. Calculate f(1).
    f(1)=13−3⋅1+1=1−3+1f(1) = 1^3 - 3\cdot 1 + 1 = 1 - 3 + 1
    f(1)=13−3⋅1+1=1−3+1=−1<0f(1) = 1^3 - 3\cdot1 + 1 = 1 - 3 + 1 = \mathbf{-1} < 0. The function is negative at the left endpoint.
  2. Calculate f(2).
    f(2)=23−3⋅2+1=8−6+1f(2) = 2^3 - 3\cdot 2 + 1 = 8 - 6 + 1
    f(2)=23−3⋅2+1=8−6+1=3>0f(2) = 2^3 - 3\cdot2 + 1 = 8 - 6 + 1 = \mathbf{3} > 0. The function is positive at the right endpoint.
  3. Do f(1) and f(2) have opposite signs? What does Bolzano say?
    f(1)⋅f(2)=−1⋅3=−3<0f(1) \cdot f(2) = -1 \cdot 3 = -3 < 0
    f(1)=−1<0f(1) = -1 < 0 and f(2)=3>0f(2) = 3 > 0 have opposite signs. Since ff is continuous on [1,2][1,2] (it is a polynomial), by Bolzano's theorem there exists c∈(1,2)c \in (1,2) such that f(c)=0f(c) = \mathbf{0}.
Result:By Bolzano's theorem, there exists c ∈ (1, 2) such that f(c) = 0.
TheoryContinuitàFormula sheetCalculus 1

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