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Posted by Manisha Behera 3 years ago (4838801)
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Preeti Dabral 3 years ago (2983787)
We have
f (x) = 2x3 – 6x2 + 6x + 5
or f ′(x) = 6x2 – 12x + 6 = 6 (x – 1)2
Now, f ′(x) = 0
{tex}\Rightarrow{/tex} x = 1
Thus, x = 1 is the only critical point of f . We shall now examine this point for local maxima and/or local minima of f. Observe that f ′(x) {tex}\ge{/tex} 0, for all x {tex}\in{/tex} R and in particular f ′(x) > 0, for values close to 1, to the left and to the right of 1. Therefore, by first derivative test, the point x = 1 is neither a point of local maxima nor a point of local minima. Hence x = 1 is a point of inflexion.
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Noobarta Meher 3 years, 3 months ago (13460227)
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