The function f(x) = 4 - 3x + 3x^2 - x^3 is (Here mathbbR is set of real numbers)

To check if a function is increasing or decreasing, find its derivative and check if it's positive or negative.

If $f'(x) > 0$f′(x)>0, the function is increasing.
If $f'(x) < 0$f′(x)<0, the function is decreasing.

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Given: $f(x) = 4 - 3x + 3x^2 - x^3$f(x)=4−3x+3x2−x3

Using the power rule:

$$f'(x) = 0 - 3(1) + 3(2x) - 3x^2$$f′(x)=0−3(1)+3(2x)−3x2

$$f'(x) = -3 + 6x - 3x^2$$f′(x)=−3+6x−3x2

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Factoring out the common factor $-3$−3:

$$f'(x) = -3(1 - 2x + x^2)$$f′(x)=−3(1−2x+x2)

The expression $1 - 2x + x^2$1−2x+x2 is a perfect square since $(x - 1)^2 = x^2 - 2x + 1$(x−1)2=x2−2x+1

Therefore:

$$f'(x) = -3(x - 1)^2$$f′(x)=−3(x−1)2

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The term $(x - 1)^2$(x−1)2 is always $\geq 0$≥0 for any value of $x$x.

Therefore:

$-3(x - 1)^2 \leq 0$−3(x−1)2≤0 for all $x$x

This means $f'(x) \leq 0$f′(x)≤0 for all real numbers.

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Since $f'(x) \leq 0$f′(x)≤0 for all $x \in \mathbb{R}$x∈R, the function is decreasing on $\mathbb{R}$R.

Note: Although $f'(x) = 0$f′(x)=0 at $x = 1$x=1, this is just one point. The function is still considered decreasing on all of $\mathbb{R}$R.

Therefore, the function is decreasing on $\mathbb{R}$R.