Chapter 3.6, Problem 104PE

(a)

To determine

Let a, b, and c are positive real numbers, where $a<b<c$ and $g\left(x\right)=\frac{\left(a-x\right){\left(x-b\right)}^2}{{\left(c-x\right)}^5}$. Then the missing values in the sign chart.

Sign of $\left(a-x\right)$.
Sign of ${\left(x-b\right)}^2$.
Sign of ${\left(c-x\right)}^5$.
Sign of $\frac{\left(a-x\right){\left(x-b\right)}^2}{{\left(c-x\right)}^5}$.

a bc

(b)

To determine

Let a, b, and c are positive real numbers, where $a<b<c$ and $g\left(x\right)=\frac{\left(a-x\right){\left(x-b\right)}^2}{{\left(c-x\right)}^5}$. Then the solution set for $g\left(x\right)>0$.

Sign of $\left(a-x\right)$.	+	$-$	$-$	$-$
Sign of ${\left(x-b\right)}^2$.	+	+	+	+
Sign of ${\left(c-x\right)}^5$.	+	+	+	$-$
Sign of $\frac{\left(a-x\right){\left(x-b\right)}^2}{{\left(c-x\right)}^5}$.	+	$-$	$-$	+

a bc

(c)

To determine

The solution set for $g\left(x\right)<0$, if a,b, and c are positive real numbers, where $a<b<c$ and $g\left(x\right)=\frac{\left(a-x\right){\left(x-b\right)}^2}{{\left(c-x\right)}^5}$.

Sign of $\left(a-x\right)$.	+	$-$	$-$	$-$
Sign of ${\left(x-b\right)}^2$.	+	+	+	+
Sign of ${\left(c-x\right)}^5$.	+	+	+	$-$
Sign of $\frac{\left(a-x\right){\left(x-b\right)}^2}{{\left(c-x\right)}^5}$.	+	$-$	$-$	+

a bc