Find the critical numbers, the intervals on which $$ f(x) $$ is increasing, The intervals on which $$ f(x) $$ is decreasing, and the local extrema. Do not graph. $$ f(x)=x+\frac{9}{x} $$ Select the correct choice helow and, if necessary, fill in the answer bax to cornplete your choice. A. The function is decreasing on $$ \square $$ I. (Type your answer in interval notation. Type integers or simplified fractions. Use a cumma to separale answers as noeded.) B. The function is never decreasing. Seled the correct choice below and, if necessary, fil in the answer boxes to complete your choice. (Type integers or simplified fractions.) A. The function has a local minimum if $$ (\square)=\square $$, and no local maximum. B. The function has a local maximum I( $$ \square $$ $$ \square $$ C. The function has a local maximum $$ f(\square)=\square $$, and no local minimum D. The function has no local extrema.

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**Step 1. Find the derivative and the critical numbers** The function is $$ f(x)=x+\frac{9}{x}. $$ Differentiate with respect to $$x$$: $$ f'(x)=\frac{d}{dx}\left(x+\frac{9}{x}\right)=1-\frac{9}{x^2}. $$ Critical numbers occur when $$f'(x)=0$$ or when $$f'(x)$$ is undefined (provided the point is in the domain). Set the derivative equal to zero: $$ 1-\frac{9}{x^2}=0 \quad \Longrightarrow \quad \frac{9}{x^2}=1 \quad \Longrightarrow \quad x^2=9. $$ Thus, $$ x=3\quad \text{or} \quad x=-3. $$ Also note $$f'(x)$$ is undefined at $$x=0$$, but $$x=0$$ is not in the domain of $$f(x)$$. **Step 2. Determine the intervals on which $$f(x)$$ is increasing or decreasing** We analyze the sign of $$f'(x)=1-\frac{9}{x^2}$$ in the intervals determined by the critical numbers and the point $$x=0$$ (which is excluded). This gives us four intervals: $$ (-\infty,-3),\quad (-3,0),\quad (0,3),\quad (3,\infty). $$ - For $$x \in (-\infty,-3)$$: Choose $$x=-4$$. Then, $$ f'(-4)=1-\frac{9}{16}>0. $$ So, $$f(x)$$ is increasing on $$(-\infty,-3)$$. - For $$x \in (-3,0)$$: Choose $$x=-1$$. Then, $$ f'(-1)=1-\frac{9}{1}=1-9=-8<0. $$ So, $$f(x)$$ is decreasing on $$(-3,0)$$. - For $$x \in (0,3)$$: Choose $$x=1$$. Then, $$ f'(1)=1-9=-8<0. $$ So, $$f(x)$$ is decreasing on $$(0,3)$$. - For $$x \in (3,\infty)$$: Choose $$x=4$$. Then, $$ f'(4)=1-\frac{9}{16}>0. $$ So, $$f(x)$$ is increasing on $$(3,\infty)$$. Thus, the function is decreasing on $$ (-3, 0)\cup(0, 3). $$ **Step 3. Determine the local extrema** - At $$x=-3$$: The sign of $$f'(x)$$ changes from positive (in $$(-\infty,-3)$$) to negative (in $$(-3,0)$$). Therefore, $$x=-3$$ is a local maximum. Compute: $$ f(-3)=-3+\frac{9}{-3}=-3-3=-6. $$ - At $$x=3$$: The sign of $$f'(x)$$ changes from negative (in $$(0,3)$$) to positive (in $$(3,\infty)$$). Therefore, $$x=3$$ is a local minimum. Compute: $$ f(3)=3+\frac{9}{3}=3+3=6. $$ Thus, the function has a local maximum at $$x=-3$$ with value $$-6$$ and a local minimum at $$x=3$$ with value $$6$$. **Answers:** 1. **Critical Numbers:** $$x=-3,\, 3$$. 2. **Intervals on which $$f(x)$$ is increasing:** $$(-\infty,-3)$$ and $$(3,\infty)$$. 3. **Intervals on which $$f(x)$$ is decreasing:** $$(-3,0)\cup(0,3)$$. 4. **Local Extrema:** The function has a local maximum at $$x=-3$$ with $$f(-3)=-6$$ and a local minimum at $$x=3$$ with $$f(3)=6$$. In the given multiple choice format: - For the first part (decreasing intervals), the answer is: $$(-3,0)\cup(0,3).$$ - For the local extrema, the correct choice is that the function has a local maximum at $$x=-3$$ (with value $$-6$$) and a local minimum at $$x=3$$ (with value $$6$$). **Critical Numbers:** $$x = -3$$ and $$x = 3$$. **Intervals on which $$f(x)$$ is increasing:** $$(-\infty, -3)$$ and $$(3, \infty)$$. **Intervals on which $$f(x)$$ is decreasing:** $$(-3, 0)$$ and $$(0, 3)$$. **Local Extrema:** - Local maximum at $$x = -3$$ with $$f(-3) = -6$$. - Local minimum at $$x = 3$$ with $$f(3) = 6$$.

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