Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ y=f(x) $$. $$ f(x)=18 x(x-1)^{3} $$ On what interval(s) is $$ f $$ increasing or decreasing? (Type your answer in interval notation. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) A. $$ f $$ is increasing on $$ \square $$ and decreasing on $$ \square $$ B. $$ f $$ is never decreasing; $$ f $$ is increasing on $$ \square $$ C. $$ f $$ is never increasing; $$ f $$ is decreasing on $$ \square $$ On what interval(s) is $$ f $$ concave upward or concave downward? (Type your answer in interval notation. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) A. $$ f $$ is concave upward on $$ \square $$ and concave downward on $$ \square $$ B. $$ f $$ is never concave downward; $$ f $$ is concave upward on $$ \square $$ C. $$ f $$ is never concave upward; $$ f $$ is concave downward on $$ \square $$

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We begin with the function $$ f(x)=18x(x-1)^3. $$ ### Step 1. Zeros and End Behavior The zeros occur when either factor is zero: - $$x=0$$, - $$x-1=0$$ so $$x=1$$ (with multiplicity 3). Since the highest degree term comes from $$ 18x \cdot (x)^3 = 18x^4, $$ the end behavior is that $$f(x)\to +\infty$$ as $$x\to \pm\infty$$. ### Step 2. First Derivative: Increasing and Decreasing Intervals Differentiate using the product rule. Let $$ f(x)=18 \cdot u(x), \quad \text{where} \quad u(x)=x(x-1)^3. $$ Differentiate $$u(x)$$: $$ u'(x)=(x-1)^3+x\cdot3(x-1)^2. $$ Factor out $$(x-1)^2$$: $$ u'(x)=(x-1)^2\big[(x-1)+3x\big]=(x-1)^2(4x-1). $$ So, $$ f'(x)=18(x-1)^2(4x-1). $$ Set $$f'(x)=0$$: - $$(x-1)^2=0$$ gives $$x=1$$, - $$4x-1=0$$ gives $$x=\frac{1}{4}$$. Since $$(x-1)^2$$ is always nonnegative, the sign of $$f'(x)$$ is determined by $$4x-1$$: - For $$x<\frac{1}{4}$$: $$4x-1<0$$ so $$f'(x)<0$$ (function is decreasing), - For $$x>\frac{1}{4}$$: $$4x-1>0$$ so $$f'(x)>0$$ (function is increasing). Note that at $$x=1$$ the derivative is zero but the sign does not change (it remains positive on both sides), so $$x=1$$ is not a local extreme point. Thus, we have: - $$f$$ is decreasing on $$(-\infty, \frac{1}{4})$$, - $$f$$ is increasing on $$(\frac{1}{4}, \infty)$$. ### Step 3. Second Derivative: Concavity Differentiate $$f'(x)=18(x-1)^2(4x-1)$$. Let $$ u(x)=(x-1)^2 \quad \text{and} \quad v(x)=4x-1. $$ Then $$ u'(x)=2(x-1) \quad \text{and} \quad v'(x)=4. $$ Using the product rule: $$ f''(x)=18\Big[u'(x)v(x)+u(x)v'(x)\Big]=18\Big[2(x-1)(4x-1)+4(x-1)^2\Big]. $$ Factor out $$(x-1)$$: $$ f''(x)=18(x-1)\Big[2(4x-1)+4(x-1)\Big]. $$ Simplify inside the bracket: $$ 2(4x-1)+4(x-1)=8x-2+4x-4=12x-6=6(2x-1). $$ Thus, $$ f''(x)=18(x-1)\cdot6(2x-1)=108(x-1)(2x-1). $$ Set $$f''(x)=0$$: - $$x-1=0$$ gives $$x=1$$, - $$2x-1=0$$ gives $$x=\frac{1}{2}$$. Test the sign of $$f''(x)$$ in the intervals determined by $$x=\frac{1}{2}$$ and $$x=1$$: 1. For $$x<\frac{1}{2}$$ (e.g., $$x=0$$): - $$x-1<0$$ and $$2x-1<0$$ so the product $$(x-1)(2x-1)$$ is positive. - Hence, $$f''(x)>0$$ (concave upward). 2. For $$\frac{1}{2}0$$ so the product is negative. - Hence, $$f''(x)<0$$ (concave downward). 3. For $$x>1$$ (e.g., $$x=2$$): - $$x-1>0$$ and $$2x-1>0$$ so the product is positive. - Hence, $$f''(x)>0$$ (concave upward). Thus, the concavity is: - $$f$$ is concave upward on $$(-\infty, \frac{1}{2})\cup(1,\infty)$$, - $$f$$ is concave downward on $$(\frac{1}{2}, 1)$$. ### Step 4. Summary Answers For the intervals of increasing/decreasing: - A. $$f$$ is increasing on $$\left(\frac{1}{4},\infty\right)$$ and decreasing on $$\left(-\infty,\frac{1}{4}\right)$$. For the intervals of concavity: - A. $$f$$ is concave upward on $$\left(-\infty,\frac{1}{2}\right)\cup\left(1,\infty\right)$$ and concave downward on $$\left(\frac{1}{2},1\right)$$. - **Increasing and Decreasing Intervals:** - $$ f $$ is increasing on $$ \left(\frac{1}{4}, \infty\right) $$ and decreasing on $$ \left(-\infty, \frac{1}{4}\right) $$. - **Concavity Intervals:** - $$ f $$ is concave upward on $$ \left(-\infty, \frac{1}{2}\right) $$ and $$ \left(1, \infty\right) $$. - $$ f $$ is concave downward on $$ \left(\frac{1}{2}, 1\right) $$.

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