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Question
Solve using intermediate value theorem:
Show that x3 − 5x2 + 3x + 6 = 0 has at least two real root between x = 1 and x = 5
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Solution
Let f(x) = x3 − 5x2 + 3x + 6 which is a polynomial function and hence continuous on [1, 5]
We factorise f(x) by synthetic division:
| 2 |
1 –5 3 6 2 –6 –6 |
| 1 –3 –3 0 |
∴ f(x) = (x – 2)(x2 – 3x – 3)
∴ f(1) = (1 – 2)(1 – 3 – 3) = 5 > 0
f(2) = (2 – 2)(4 – 6 – 3) = 0
∴ x = 2 is a root of f(x) = 0
f(3) = (3 – 2)(9 – 9 – 3) = – 3 < 0
f(4) = (4 – 2)(16 – 12 – 3) = 2 > 0
f is continuous on [3, 4]
f(3) < 0, f(4) > 0
∴ by intermediate value theorem for continuous function f(x) = 0 has a root between 3 and 4
∴ there are two roots, x = 2 and a root between x = 3 and x = 4.
∴ f(x) = 0 has at least two root between 1 and 5.
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