Solve the nonlinear inequality: (x − 1)(x+3)(x − 7)2 ≥ 0 Solve by marking zeros and undefined points on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.
Solve the nonlinear inequality: (x − 1)(x+3)(x − 7)2 ≥ 0 Solve by marking zeros and undefined points on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.
A product is greater zero if the number of negative signs is even.
Since $\displaystyle (x-7)^2 > 0~\implies~x \in \mathbb{R} \setminus \{7\}$ you only have to examine the remaining 2 factors:
$\displaystyle \underbrace{x-1\geq 0\wedge x+3 \geq 0}_{\text{both factors positive}}~\vee~ \underbrace{x-1< 0\wedge x+3 < 0}_{\text{both factors negative}}$
After some simplification you'll get:
$\displaystyle x \in (-\infty,-3] \cup [1,\infty)$
Solve the inequality:x(x + 2)(x − 5)(x + 7) ≤ 0 Solve by marking zeros and undefined points
on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.
Hello, cwaddy!
The zeros of the inequality are: .$\displaystyle -7,\:-2,\:0,\:5$Solve the inequality: .$\displaystyle x(x+ 2)(x - 5) (x + 7)\:\leq\: 0$
Solve by marking zeros and undened points on a number line
and checking signs in the intervals created.
Express your answer in interval notation.
These four values divide the number line into five intervals.
. . $\displaystyle \begin{array}{cccccccccc}
\underbrace{\;-----} & o & \underbrace{----- }& o & \underbrace{-----} & o & \underbrace{-----} & o & \underbrace{-----\;} \\
& \text{-}7 && \text{-}2 && 0 && 5 \end{array}$
Now test a value from each interval:
On $\displaystyle (\text{-}\infty, \text{-}7)$, try $\displaystyle x = \text{-}8:\;\;(\text{-}8)(\text{-}8+2)(\text{-}8-5)(\text{-}8+7) \:=\:(\text{-}8)(\text{-}6)(\text{-}13)(\text{-}1) \:=\:+182$ . . . no
On $\displaystyle (\text{-}7,\text{-}2)$, try $\displaystyle x = \text{-}3\!:;\;(\text{-}3)(\text{-}3+2)(\text{-}3-5)(\text{-}3+7) \:=\:(\text{-}3)(\text{-}1)(\text{-}8)(4) \:=\:-96$ . . . yes
On $\displaystyle (\text{-}2,0)$, try $\displaystyle x = \text{-}1\!:\;\;(\text{-}1)(\text{-}1+2)(\text{-}1-5)(\text{-}1+7) \:=\:(\text{-}1)(1)(\text{-}6)(6) \:=\:+36$ . . . no
On $\displaystyle (0,5)$, try $\displaystyle x = 1\!:\;\;(1)(1+7)(1-5)(1+7) \:=\:(1)(8)(\text{-}6)(6) \:=\:-288$ . . . yes
On$\displaystyle (5,\infty)$, try $\displaystyle x = 6\!:\;\;(6)(6+2)(6-5)(6+7) \:=\:(6)(8)(1)(13) \:=\:+624$ . . . no
The inequality is satisfied on: .$\displaystyle (\text{-}7,\text{-}2) \,\cup\, (0,5)$