If f (x) = x³ – 6x² + 9x + 3 be a decreasing function, then x lies in

The correct option is (1, 3).

To find the interval in which the function $f\left(x\right)=x^3-6x^2+9x+3$ is decreasing, we need to analyze its first derivative.

Recall that a differentiable function $f\left(x\right)$ is decreasing on an interval if its derivative is less than zero for all $x$ in that interval:
$f^{\prime }\left(x\right)<0$

First, let us find the derivative of $f\left(x\right)$ with respect to $x$:
$f^{\prime }\left(x\right)=\frac{d}{dx}(x^3-6x^2+9x+3)$

Applying the power rule for differentiation, we get:
$f^{\prime }\left(x\right)=3x^2-12x+9$

For the function to be decreasing, we set $f^{\prime }\left(x\right)<0$:
$3x^2-12x+9<0$

We can simplify this inequality by dividing all terms by 3:
$x^2-4x+3<0$

Now, we factor the quadratic expression $x^2-4x+3$. We look for two numbers that multiply to 3 and add up to -4. These numbers are -1 and -3. Thus, the factored form is:
$(x-1)(x-3)<0$

To solve this inequality, we find the critical points where the expression equals zero, which are $x=1$ and $x=3$. These points divide the real number line into three intervals: $(-\infty ,1)$, $(1,3)$, and $(3,\infty )$.

We test the sign of the product $(x-1)(x-3)$ in each interval:
1. For $x<1$ (e.g., $x=0$): both terms are negative, so their product is positive: $(-)\times (-)>0$.
2. For $1<x<3$ (e.g., $x=2$): the first term is positive and the second term is negative, so their product is negative: $(+)\times (-)<0$.
3. For $x>3$ (e.g., $x=4$): both terms are positive, so their product is positive: $(+)\times (+)>0$.

Therefore, the inequality $(x-1)(x-3)<0$ is satisfied when $x$ lies in the interval:
$x\in (1,3)$