The number of real roots of the eqation $2 x^{3} - 3 x^{2} + 6 x + 6 = 0$ is

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Solution

The correct option is A 1
$f (x) = 2 x^{3} - 3 x^{2} + 6 x + 6$
$f^{'} (x) = 6 x^{2} - 6 x + 6$
$f^{'} (x) > 0$ (always) $f (x)$ is monotonically increasing.
So this equation have two imaginary and one real root.

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The number of real roots of the eqation 2x3−3x2+6x+6=0 is

The correct option is A 1f(x)=2x3−3x2+6x+6f′(x)=6x2−6x+6f′(x)>0(always) f(x) is monotonically increasing.So this equation have two imaginary and one real root.

The number of real roots of the eqation $2 x^{3} - 3 x^{2} + 6 x + 6 = 0$ is

The correct option is A 1
$f (x) = 2 x^{3} - 3 x^{2} + 6 x + 6$
$f^{'} (x) = 6 x^{2} - 6 x + 6$
$f^{'} (x) > 0$ (always) $f (x)$ is monotonically increasing.
So this equation have two imaginary and one real root.