Math · Calculus I · Worked example
Trap a root with the Intermediate Value Theorem
Show that x³ + x − 1 = 0 has a solution between 0 and 1, then narrow it down by bisection.
Check continuity
f(x) = x³ + x − 1 is a polynomial, so it is continuous on [0, 1].
Evaluate the endpoints
f(0) = −1 < 0 and f(1) = 1 > 0. Zero lies between them, so f(c) = 0 for some c in (0, 1).
Halve the interval
f(0.5) = −0.375 < 0, so the sign change, and a root, is in (0.5, 1). Then f(0.75) = 0.171875 > 0 puts it in (0.5, 0.75).
Keep halving
f(0.625) ≈ −0.131 and f(0.6875) ≈ 0.0125, so the root is in (0.625, 0.6875). Each step halves the interval.
Result
A root lies in (0, 1); after four halvings it lies in (0.625, 0.6875). It is about 0.6823.
Your turn
Show that cos x = x has a solution between 0 and 1.
Show the answer and explanation
g(x) = cos x − x is continuous, g(0) = 1 > 0 and g(1) = cos 1 − 1 ≈ −0.46 < 0, so g(c) = 0 for some c in (0, 1).
Rewrite the equation as g(x) = 0. The Intermediate Value Theorem applies because g is continuous and changes sign on [0, 1].
Keep exploring
In Graph, y = x³ + x − 1 crosses the x-axis once, near x = 0.682. Change −1 to +1 and the crossing moves to about −0.682; then f(−1) = −1 and f(0) = 1 bracket it.
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