Proove that there is a root between 1 and 2

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Prove that the equation [tex]4x^3-6x^2+3x-2=0[/tex] has one root exactly between 1 and 2.

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Spoiler: show

[tex]4x^3-6x^2+3x-2 = \frac{1}{2}((2x-1)^3-3)[/tex]

R.Baber.

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Let f(x)=4x^3-6x^2+3x-2
If we get different signs for the value of f(x) when we substitute x=1 and x=2 in the function, we can assume that the there lies a root of f(x) between 1 and 2
Now let us substitute x=1 in the given expression
f(1)=4-6+3-2=-1
Now substitute x=2in the given expression
f(2)=32-24+6-2=+12
Since the value of f(x) is changing when we substitute x=1 and x=2 we can deduce that one root of of the given function is lying between 1 and 2