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Question
Show that there is a root for the equation 2x3 − x − 16 = 0 between 2 and 3.
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Solution
Let f(x) = 2x3 − x − 16
f(x) is a polynomial function and hence it is continuous for all x ∈ R
A root of f(x) exists if f(x) = 0 for at least one value of x
f(2) = 2(2)3 – 2 – 16
= – 2 < 0
f(3) = 2(3)3 – 3 – 16
= 35 > 0
∴ f(2) < 0 and f(3) > 0
∴ By intermediate value theorem, there has to be point ‘c’ between 2 and 3 such that f(c) = 0
∴ There is a root of the given equation between 2 and 3.
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