The point of inflection of the function f x - x 3-3 x 2-3 x -8 - x - R exist at which value of xA- Does not existB- -1C- -2D- 1

The correct option is B 1f(x)= x^3 + 3x^2 + 3x + 8 ⇒ f'(x) = 3x^2 + 6x + 3 = 0 ⇒ x^2 + 2 x + 1 = 0 ⇒ x = 1 f”(x) = 6x + 6 ⇒ f”( 1) = 6 + 6 = 0 So, x = 1 is ...

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Standard XII

Mathematics

Strictly Increasing Functions

The point of ...

Question

The point of inflection of the function $f (x) = x^{3} + 3 x^{2} + 3 x + 8; x \in R$ exist at which value of $x$

- 1

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Solution

The correct option is A $- 1$
$f (x) = x^{3} + 3 x^{2} + 3 x + 8 \Rightarrow f^{'} (x) = 3 x^{2} + 6 x + 3 = 0 \Rightarrow x^{2} + 2 x + 1 = 0 \Rightarrow x = - 1$
$f^{''} (x) = 6 x + 6 \Rightarrow f^{''} (- 1) = - 6 + 6 = 0$
So, $x = - 1$ is a point of inflection.

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The point of inflection of the function f(x)=x3+3x2+3x+8 ; x∈R exist at which value of x

The correct option is A −1f(x)=x3+3x2+3x+8⇒f′(x)=3x2+6x+3=0⇒x2+2x+1=0⇒x=−1 f′′(x)=6x+6⇒f′′(−1)=−6+6=0 So, x=−1 is a point of inflection.

The point of inflection of the function $f (x) = x^{3} + 3 x^{2} + 3 x + 8; x \in R$ exist at which value of $x$

The correct option is A $- 1$
$f (x) = x^{3} + 3 x^{2} + 3 x + 8 \Rightarrow f^{'} (x) = 3 x^{2} + 6 x + 3 = 0 \Rightarrow x^{2} + 2 x + 1 = 0 \Rightarrow x = - 1$
$f^{''} (x) = 6 x + 6 \Rightarrow f^{''} (- 1) = - 6 + 6 = 0$
So, $x = - 1$ is a point of inflection.