28. Find the interval(s) on which $$ f(x) $$ is decreasing, if $$ f^{\prime}(x)=x^{4}-3 x^{3}+2 x^{2} $$ $$ \begin{array}{llll} ext { A. }(1,2) & ext { B. }(0, \infty) & ext { C. }(-\infty, 1) & ext { D. }(0,1),(2, \infty)\end{array} $$

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28. Find the interval(s) on which $$ f(x) $$ is decreasing, if $$ f^{\prime}(x)=x^{4}-3 x^{3}+2 x^{2} $$ $$ \begin{array}{llll}	ext { A. }(1,2) & 	ext { B. }(0, \infty) & 	ext { C. }(-\infty, 1) & 	ext { D. }(0,1),(2, \infty)\end{array} $$

UpStudy AI · Accepted Answer

To determine the intervals where the function $$ f(x) $$ is decreasing, we need to analyze the sign of its derivative $$ f'(x) $$. Specifically, $$ f(x) $$ is decreasing where $$ f'(x) < 0 $$.

Given:
$$ f'(x) = x^4 - 3x^3 + 2x^2 $$

**Step 1: Factor the Derivative**
Factor the expression to find the critical points:
$$
f'(x) = x^2(x^2 - 3x + 2) = x^2(x - 1)(x - 2)
$$

**Step 2: Analyze the Sign**
- The term $$ x^2 $$ is always non-negative ($$ x^2 \geq 0 $$) and equals zero at $$ x = 0 $$.
- The sign of $$ f'(x) $$ is primarily determined by the product $$ (x - 1)(x - 2) $$.

Evaluate $$ (x - 1)(x - 2) $$:
- **For $$ x < 1 $$**: Both $$ (x - 1) $$ and $$ (x - 2) $$ are negative, so their product is positive.
- **For $$ 1 < x < 2 $$**: $$ (x - 1) $$ is positive and $$ (x - 2) $$ is negative, so their product is negative.
- **For $$ x > 2 $$**: Both $$ (x - 1) $$ and $$ (x - 2) $$ are positive, so their product is positive.

**Step 3: Determine Where $$ f'(x) < 0 $$**
$$ f'(x) < 0 $$ only when $$ (x - 1)(x - 2) < 0 $$, which occurs in the interval $$ (1, 2) $$.

**Conclusion:**
The function $$ f(x) $$ is decreasing on the interval $$ (1, 2) $$.

**Answer:** A. $$ (1,2) $$
$$ f(x) $$ is decreasing on the interval $$ (1, 2) $$.

Find the interval(s) on which is decreasing, if

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