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12.3: Exercises

Last updated

May 2, 2022
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- 12.2: Rational inequalities and absolute value inequalities
- 13: Exponential and Logarithmic Functions

Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

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Exercise $\PageIndex{1}$

Solve for $x$ .

$5x+6\leq 21$
$3+4x> 10x$
$2x+8\geq 6x+24$
$9-3x< 2x-13$
$-5 \leq 2x+5\leq 19$
$15> 7-2x\geq 1$
$3x+4 \leq 6x-2\leq 8x+5$
$5x+2< 4x-18\leq 7x+11$

Answer

$x \leq 3$
$\dfrac{1}{2}>x$
$-4 \geq x$
$x>\dfrac{22}{5}$
$-5 \leq x \leq 7$
$-4<x \leq 3$
$x \geq 2$ (this then also implies $x \geq-\dfrac{7}{2}$ )
no solution

Exercise $\PageIndex{2}$

Solve for $x$ .

$x^2-5x-14>0$
$x^2-2x\geq 35$
$x^2-4\leq 0$
$x^2+3x-3<0$
$2x^2+2x\leq 12$
$3x^2<2x+1$
$x^2-4x+4>0$
$x^3-2x^2-5x+6\geq 0$
$x^3+4x^2+3x+12 <0$
$-x^3-4x<-4x^2$
$x^4-10x^2+9\leq 0$
$x^4-5x^3+5x^2+5x<6$
$x^4-5x^3+6x^2>0$
$x^5-6x^4+x^3+24x^2-20x\leq 0$
$x^5-15x^4+85x^3-225x^2+274x-120\geq 0$
$x^{11}-x^{10}+x-1\leq 0$

Answer

$(-\infty,-2) \cup(7, \infty)$
$(-\infty,-5] \cup[7, \infty)$
$[-2,2]$
$\left(\dfrac{-3-\sqrt{21}}{2}, \dfrac{-3+\sqrt{21}}{2}\right)$
$[-3,2]$
$\left(-\dfrac{1}{3}, 1\right)$
$\mathbb{R}-\{2\}$
$[-2,1] \cup[3, \infty)$
$(-\infty,-4)$
$(0,2) \cup(2, \infty)$
$[-3,-1] \cup[1,3]$
$(-1,1) \cup(2,3)$
$(-\infty, 0) \cup(0,2) \cup(3, \infty)$
$(-\infty,-2] \cup[0,1] \cup[2,5]$
$[1,2] \cup[3,4] \cup [5, \infty)$
$(-\infty, 1]$

Exercise $\PageIndex{3}$

Find the domain of the functions below.

$f(x)=\sqrt{x^2-8x+15}$
$f(x)=\sqrt{9x-x^3}$
$f(x)=\sqrt{(x-1)(4-x)}$
$f(x)=\sqrt{(x-2)(x-5)(x-6)}$
$f(x)=\dfrac{5}{\sqrt{6-2x}}$
$f(x)=\dfrac{1}{\sqrt{x^2-6x-7}}$

Answer

$D=(-\infty, 3] \cup[5, \infty)$
$D=(-\infty,-3] \cup[0,3]$
$D=[1,4]$
$D=[2,5] \cup[6, \infty)$
$D=(-\infty, 3)$
$D=(-\infty,-1) \cup(7, \infty)$

Exercise $\PageIndex{4}$

Solve for $x$ .

$\dfrac{x-5}{2-x}>0$
$\dfrac{4x-4}{x^2-4}\geq 0$
$\dfrac{x-2}{x^2-4x-5}< 0$
$\dfrac{x^2-9}{x^2-4}\geq 0$
$\dfrac{x-3}{x+3}\leq 4$
$\dfrac{1}{x+10}> 5$
$\dfrac{2}{x-2}\leq \dfrac{5}{x+1}$
$\dfrac{x^2}{x+4}\leq x$

Answer

$(2,5)$
$(-2,1] \cup(2, \infty)$
$(-\infty,-1) \cup(2,5)$
$(-\infty,-3] \cup (-2,2) \cup[3, \infty)$
$(-\infty,-5] \cup(-3, \infty)$
$(-10,-9.8)$
$(-1,2) \cup [4, \infty)$
$(-\infty,-4) \cup[0, \infty)$

Exercise $\PageIndex{5}$

Solve for $x$ .

$|2x+7|>9$
$|6x+2|<3$
$|5-3x|\geq 4$
$|-x-7|\leq 5$
$|1-8x|\geq 3$
$1>\left|2+\dfrac{x}{5}\right|$

Answer

$(-\infty,-8) \cup(1, \infty)$
$\left(\dfrac{-5}{6}, \dfrac{1}{6}\right)$
$\left(-\infty, \dfrac{1}{3}\right] \cup[3, \infty)$
$[-12,-2]$
$\left(-\infty,-\dfrac{1}{4}\right] \cup\left[\dfrac{1}{2}, \infty\right)$
$(-15,-5)$

Exercise 12.3.1\PageIndex{1}

Exercise 12.3.2\PageIndex{2}

Exercise 12.3.3\PageIndex{3}

Exercise 12.3.4\PageIndex{4}

Exercise 12.3.5\PageIndex{5}

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Exercise $\PageIndex{1}$

Exercise $\PageIndex{2}$

Exercise $\PageIndex{3}$

Exercise $\PageIndex{4}$

Exercise $\PageIndex{5}$