1.

Show that there is a root for the equation 2x3 – x – 16 = 0 between 2 and 3.

Answer»

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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