Give the domain of each function defined as follows. $$ \begin{array}{ll} ext { 17. } f(x)=2 x & ext { 18. } f(x)=2 x+3 \ ext { 19. } f(x)=x^{4} & ext { 20. } f(x)=(x+3)^{2} \ ext { 21. } f(x)=\sqrt{4-x^{2}} & ext { 22. } f(x)=|3 x-6| \ ext { 23. } f(x)=(x-3)^{1 / 2} & ext { 24. } f(x)=(3 x+5)^{1 / 2} \ ext { 25. } f(x)=\frac{2}{1-x^{2}} & ext { 26. } f(x)=\frac{-8}{x^{2}-36} \ ext { 27. } f(x)=-\sqrt{\frac{2}{x^{2}-16}} & ext { 28. } f(x)=-\sqrt{\frac{1}{x^{2}+36}} \ ext { 29. } f(x)=\sqrt{x^{2}-4 x-5} & ext { 30. } f(x)=\sqrt{15 x^{2}+x-2} \ ext { 31. } f(x)=\frac{1}{\sqrt{3 x^{2}+2 x-1}} & \end{array} $$

Question

Give the domain of each function defined as follows. $$ \begin{array}{ll}	ext { 17. } f(x)=2 x & 	ext { 18. } f(x)=2 x+3 \ 	ext { 19. } f(x)=x^{4} & 	ext { 20. } f(x)=(x+3)^{2} \ 	ext { 21. } f(x)=\sqrt{4-x^{2}} & 	ext { 22. } f(x)=|3 x-6| \ 	ext { 23. } f(x)=(x-3)^{1 / 2} & 	ext { 24. } f(x)=(3 x+5)^{1 / 2} \ 	ext { 25. } f(x)=\frac{2}{1-x^{2}} & 	ext { 26. } f(x)=\frac{-8}{x^{2}-36} \ 	ext { 27. } f(x)=-\sqrt{\frac{2}{x^{2}-16}} & 	ext { 28. } f(x)=-\sqrt{\frac{1}{x^{2}+36}} \ 	ext { 29. } f(x)=\sqrt{x^{2}-4 x-5} & 	ext { 30. } f(x)=\sqrt{15 x^{2}+x-2} \ 	ext { 31. } f(x)=\frac{1}{\sqrt{3 x^{2}+2 x-1}} & \end{array} $$

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**17.** $$ f(x)=2x $$ All real numbers. Domain: $$\mathbb{R}$$. **18.** $$ f(x)=2x+3 $$ All real numbers. Domain: $$\mathbb{R}$$. **19.** $$ f(x)=x^4 $$ All real numbers. Domain: $$\mathbb{R}$$. **20.** $$ f(x)=(x+3)^2 $$ All real numbers. Domain: $$\mathbb{R}$$. **21.** $$ f(x)=\sqrt{4-x^2} $$ For the square root to be defined, $$ 4-x^2\ge0 \implies x^2\le4 \implies -2\le x\le2. $$ Domain: $$[ -2,\,2 ]$$. **22.** $$ f(x)=|3x-6| $$ Absolute value is defined for all real numbers. Domain: $$\mathbb{R}$$. **23.** $$ f(x)=(x-3)^{1/2} $$ Interpreted as a square root, we require $$ x-3\ge0 \implies x\ge3. $$ Domain: $$[ 3,\,\infty )$$. **24.** $$ f(x)=(3x+5)^{1/2} $$ Require $$ 3x+5\ge0 \implies x\ge -\frac{5}{3}. $$ Domain: $$\left[-\frac{5}{3},\,\infty\right)$$. **25.** $$ f(x)=\frac{2}{1-x^2} $$ The denominator must be nonzero: $$ 1-x^2\neq0 \implies x^2\neq1 \implies x\neq1 \text{ and } x\neq -1. $$ Domain: $$\mathbb{R}\setminus\{-1,\,1\}$$. **26.** $$ f(x)=\frac{-8}{x^2-36} $$ Here, $$ x^2-36\neq0 \implies x^2\neq36 \implies x\neq6 \text{ and } x\neq -6. $$ Domain: $$\mathbb{R}\setminus\{-6,\,6\}$$. **27.** $$ f(x)=-\sqrt{\frac{2}{x^2-16}} $$ For the square root, the radicand must be nonnegative and defined. Since $$ \frac{2}{x^2-16} \ge0, $$ and $$2>0$$, the sign depends on the denominator. Also, $$x^2-16$$ must be nonzero (to avoid division by zero). Solve: $$ x^2-16>0 \implies x^2>16 \implies |x|>4. $$ Thus, Domain: $$(-\infty, -4)\cup(4, \infty)$$. **28.** $$ f(x)=-\sqrt{\frac{1}{x^2+36}} $$ Here, $$x^2+36>0$$ for all real $$x$$, so the expression under the square root is always positive. Domain: $$\mathbb{R}$$. **29.** $$ f(x)=\sqrt{x^2-4x-5} $$ The radicand must satisfy: $$ x^2-4x-5\ge 0. $$ Factor (or use the quadratic formula): The roots are found by solving $$ x^2-4x-5=0 \implies x=\frac{4\pm\sqrt{16+20}}{2}=\frac{4\pm6}{2}. $$ Thus, $$x=-1$$ and $$x=5$$. Test intervals or use the fact that the coefficient is positive: $$ x^2-4x-5\ge0 \quad \text{if} \quad x\le -1 \quad \text{or} \quad x\ge5. $$ Domain: $$(-\infty,-1]\cup [5,\infty)$$. **30.** $$ f(x)=\sqrt{15x^2+x-2} $$ Set the radicand nonnegative: $$ 15x^2+x-2\ge0. $$ Find the roots: $$ x=\frac{-1\pm\sqrt{1-4(15)(-2)}}{2\cdot15}=\frac{-1\pm\sqrt{1+120}}{30}=\frac{-1\pm 11}{30}. $$ This yields $$ x=\frac{10}{30}=\frac{1}{3} \quad \text{and} \quad x=\frac{-12}{30}=-\frac{2}{5}. $$ Since the leading coefficient is positive, the quadratic is nonnegative when $$ x\le -\frac{2}{5} \quad \text{or} \quad x\ge \frac{1}{3}. $$ Domain: $$\left(-\infty,-\frac{2}{5}\right]\cup\left[\frac{1}{3},\infty\right)$$. **31.** $$ f(x)=\frac{1}{\sqrt{3x^2+2x-1}} $$ For the expression to be defined, the radicand must be strictly positive (since it is in the denominator): $$ 3x^2+2x-1>0. $$ Solve the quadratic equation: $$ 3x^2+2x-1=0. $$ Using the quadratic formula: $$ x=\frac{-2\pm\sqrt{4+12}}{6}=\frac{-2\pm\sqrt{16}}{6}=\frac{-2\pm4}{6}. $$ Thus, the roots are: $$ x=\frac{1}{3} \quad \text{and} \quad x=-1. $$ Since the coefficient of $$x^2$$ is positive, the quadratic is positive when $$ x<-1 \quad \text{or} \quad x>\frac{1}{3}. $$ Domain: $$(-\infty,-1)\cup\left(\frac{1}{3},\infty\right)$$. **17.** All real numbers. Domain: $$\mathbb{R}$$. **18.** All real numbers. Domain: $$\mathbb{R}$$. **19.** All real numbers. Domain: $$\mathbb{R}$$. **20.** All real numbers. Domain: $$\mathbb{R}$$. **21.** $$ -2 \leq x \leq 2 $$ Domain: $$[ -2,\,2 ]$$. **22.** All real numbers. Domain: $$\mathbb{R}$$. **23.** $$ x \geq 3 $$ Domain: $$[ 3,\,\infty )$$. **24.** $$ x \geq -\frac{5}{3} $$ Domain: $$\left[-\frac{5}{3},\,\infty\right)$$. **25.** $$ x \neq 1 $$ and $$ x \neq -1 $$ Domain: $$\mathbb{R}\setminus\{-1,\,1\}$$. **26.** $$ x \neq 6 $$ and $$ x \neq -6 $$ Domain: $$\mathbb{R}\setminus\{-6,\,6\}$$. **27.** $$ x < -4 $$ or $$ x > 4 $$ Domain: $$(-\infty, -4)\cup(4, \infty)$$. **28.** All real numbers. Domain: $$\mathbb{R}$$. **29.** $$ x \leq -1 $$ or $$ x \geq 5 $$ Domain: $$(-\infty,-1]\cup [5,\infty)$$. **30.** $$ x \leq -\frac{2}{5} $$ or $$ x \geq \frac{1}{3} $$ Domain: $$\left(-\infty,-\frac{2}{5}\right]\cup\left[\frac{1}{3},\infty\right)$$. **31.** $$ x < -1 $$ or $$ x > \frac{1}{3} $$ Domain: $$(-\infty,-1)\cup\left(\frac{1}{3},\infty\right)$$.

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