Opening the book…
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
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
-
- \((-\infty,-2) \cup(7, \infty)\)
- \((-\infty,-5] \cup[7, \infty)\)
- \([-2,2]\)
- \(\left(\dfrac{-3-\sqrt{21}}{2}, \dfrac{-3+\sqrt{21}}{2}\right)\)
- \([-3,2]\)
- \(\left(-\dfrac{1}{3}, 1\right)\)
- \(\mathbb{R}-\{2\}\)
- \([-2,1] \cup[3, \infty)\)
- \((-\infty,-4)\)
- \((0,2) \cup(2, \infty)\)
- \([-3,-1] \cup[1,3]\)
- \((-1,1) \cup(2,3)\)
- \((-\infty, 0) \cup(0,2) \cup(3, \infty)\)
- \((-\infty,-2] \cup[0,1] \cup[2,5]\)
- \([1,2] \cup[3,4] \cup [5, \infty)\)
- \((-\infty, 1]\)
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
-
- \(D=(-\infty, 3] \cup[5, \infty)\)
- \(D=(-\infty,-3] \cup[0,3]\)
- \(D=[1,4]\)
- \(D=[2,5] \cup[6, \infty)\)
- \(D=(-\infty, 3)\)
- \(D=(-\infty,-1) \cup(7, \infty)\)
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
-
- \((2,5)\)
- \((-2,1] \cup(2, \infty)\)
- \((-\infty,-1) \cup(2,5)\)
- \((-\infty,-3] \cup (-2,2) \cup[3, \infty)\)
- \((-\infty,-5] \cup(-3, \infty)\)
- \((-10,-9.8)\)
- \((-1,2) \cup [4, \infty)\)
- \((-\infty,-4) \cup[0, \infty)\)
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
-
- \((-\infty,-8) \cup(1, \infty)\)
- \(\left(\dfrac{-5}{6}, \dfrac{1}{6}\right)\)
- \(\left(-\infty, \dfrac{1}{3}\right] \cup[3, \infty)\)
- \([-12,-2]\)
- \(\left(-\infty,-\dfrac{1}{4}\right] \cup\left[\dfrac{1}{2}, \infty\right)\)
- \((-15,-5)\)
This page titled 12.3: Exercises was last modified on Mon, 02 May 2022 15:49:47 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform.
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