possible For the function $$ h(x)=\frac{7 x}{(x+2)(x-9)} $$, solve the following inequality. $$ h(x)

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possible For the function $$ h(x)=\frac{7 x}{(x+2)(x-9)} $$, solve the following inequality. $$ h(x)<0 $$

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We are given $$ h(x)=\frac{7x}{(x+2)(x-9)} $$ and need to solve $$ h(x)<0. $$ **Step 1. Find Critical Points** Determine the zeros of the numerator and the points where the denominator is zero (the function is undefined there). - The numerator $$7x$$ is zero when $$ x=0. $$ - The denominator $$(x+2)(x-9)$$ is zero when $$ x+2=0 \quad \text{or} \quad x-9=0, $$ which gives $$ x=-2 \quad \text{or} \quad x=9. $$ Thus, the critical points are $$x=-2$$, $$x=0$$, and $$x=9$$. **Step 2. Determine the Sign on the Intervals** The critical points split the real line into four intervals: $$ (-\infty,-2),\quad (-2,0),\quad (0,9),\quad (9,\infty). $$ Test a point from each interval to determine the sign of $$h(x)$$. - **Interval $$(-\infty,-2)$$:** Choose $$x=-3$$. $$ \text{Numerator: } 7(-3) = -21 \quad (\text{negative}). $$ $$ \text{Denominator: } (-3+2)(-3-9)=(-1)(-12)=12 \quad (\text{positive}). $$ Therefore, $$h(-3)=\frac{-21}{12}<0$$. - **Interval $$(-2,0)$$:** Choose $$x=-1$$. $$ \text{Numerator: } 7(-1)=-7 \quad (\text{negative}). $$ $$ \text{Denominator: } (-1+2)(-1-9)=(1)(-10)=-10 \quad (\text{negative}). $$ Negative divided by negative is positive, so $$h(-1)>0$$. - **Interval $$(0,9)$$:** Choose $$x=1$$. $$ \text{Numerator: } 7(1)=7 \quad (\text{positive}). $$ $$ \text{Denominator: } (1+2)(1-9)=(3)(-8)=-24 \quad (\text{negative}). $$ Positive divided by negative gives a negative result, so $$h(1)<0$$. - **Interval $$(9, \infty)$$:** Choose $$x=10$$. $$ \text{Numerator: } 7(10)=70 \quad (\text{positive}). $$ $$ \text{Denominator: } (10+2)(10-9)=(12)(1)=12 \quad (\text{positive}). $$ Thus, $$h(10)=\frac{70}{12}>0$$. **Step 3. Write the Solution** We want $$h(x)<0$$. Based on our tests, the inequality holds in the intervals: - $$(-\infty,-2)$$ and - $$(0,9)$$. Recall that the points $$x=-2$$ and $$x=9$$ are not in the domain (they make the denominator zero), and at $$x=0$$, $$h(0)=0$$ (which does not satisfy the strict inequality). Thus, the solution is: $$ (-\infty,-2)\cup (0,9). $$ The solution to $$ h(x) < 0 $$ is $$ x $$ in the intervals $$ (-\infty, -2) $$ and $$ (0, 9) $$.

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For the function , solve the following inequality.

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possible For the function , solve the following inequality.