Find the intervals on which $$ f(x) $$ is increasing and the intervals on which $$ f(x) $$ is decreasing. Then sketch the graph. Add horizontal tangent lines. $$ f(x)=2-27 x+9 x^{2}-x^{3} $$ Select the correct choice below and, if necessary, fill in the answer box to complete your choice A. The function is increasing on (Type your answer using interval notation. Use a comma to separate answers as needed.) B. The function is never increasing.

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UpStudy AI · Accepted Answer

We begin by finding the derivative of 
$$
f(x)=2-27x+9x^{2}-x^{3}.
$$
Taking the derivative term by term:
$$
f'(x)=0-27+18x-3x^{2}=-27+18x-3x^{2}.
$$

Next, we factor out $$-3$$:
$$
f'(x)=-3(9-6x+x^{2}).
$$
Notice that the quadratic inside can be rewritten as:
$$
9-6x+x^{2}=(x-3)^{2}.
$$
Thus, we have
$$
f'(x)=-3(x-3)^{2}.
$$

Since $$(x-3)^{2}\ge 0$$ for all $$ x $$ and the factor $$-3$$ is negative, it follows that
$$
f'(x)\le 0 \quad \text{for all } x,
$$
with equality only when
$$
(x-3)^{2}=0\quad\Longrightarrow\quad x=3.
$$

Because $$ f'(x)<0 $$ for all $$ x\ne 3 $$ and $$ f'(3)=0 $$, it implies that the function is strictly decreasing everywhere except at $$ x=3 $$ (where the tangent is horizontal).

The horizontal tangent line occurs at $$ x=3 $$. To find its $$ y $$-coordinate, we compute:
$$
f(3)=2-27(3)+9(3)^{2}-3^{3}=2-81+81-27=-25.
$$
Thus, the horizontal tangent line is given by:
$$
y=-25 \quad \text{at} \quad x=3.
$$

Since the function is never increasing, the correct choice is:

B. The function is never increasing.

Graphing $$ f(x) $$ shows a cubic curve that is strictly decreasing with a flattening (horizontal tangent) at $$ (3,-25) $$.
The function is never increasing.

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