Answer to Question #260762 in Calculus for Sabelo Xulu

Question

The functions f and g are defined by f(x) =1/(1-3x) and g(x) =log_1/3(3x-2)-log₃(x) respectively

1. Write down the sets D_{f }(ehe domain of f) and D_g (the domain of g)

2. Solve the inequality f(x) > 2 for x$\isin$ D_f

3. Solve the inequality f(x) ≥ 2 for x$\isin$ D_g

Hint: Use the change of base formula

Answer 1

1.

$$f(x)=\dfrac{1}{1-3x}$$ $$1-3x\not=0=>x\not=\dfrac{1}{3}$$ $$Df: (-\infin, \dfrac{1}{3})\cup(\dfrac{1}{3}, \infin)$$ $$g(x)=\log_{1/3}(3x-2)-\log_{3}(x)$$ $$\begin{cases} 3x-2>0 \\ x>0 \end{cases}=>x>\dfrac{2}{3}$$ $$Dg: (\dfrac{2}{3}, \infin)$$

2.

$$f(x)>2, x\in Df$$ $$\dfrac{1}{1-3x}>2, x\in(-\infin, \dfrac{1}{3})\cup(\dfrac{1}{3}, \infin)$$ $$\dfrac{1}{1-3x}-2>0$$ $$\dfrac{1-2+6x}{1-3x}>0$$ $$\dfrac{6x-1}{3x-1}<0$$ $$\dfrac{1}{6}<x<\dfrac{1}{3}$$ $$x\in(\dfrac{1}{6},\dfrac{1}{3})$$

3.

$$g(x)\geq2, x\in Dg$$ $$\log_{1/3}(3x-2)-\log_{3}(x)\geq2, x\in(\dfrac{2}{3}, \infin)$$ $$\log_{1/3}(3x-2)=-\log_{3}(3x-2)$$ $$2=\log_{3}(9)$$ $$-\log_{3}(3x-2)-\log_{3}(x)\geq\log_{3}(9)$$ $$\log_{3}(9x(3x-2))\leq0$$ $$3>1=>y(x)=\log_{3}(x)\ increases$$

Then

$$9x(3x-2)\leq1, x\in(\dfrac{2}{3}, \infin)$$ $$27x^2-18x-1\leq0$$

Let

$$27x^2-18x-1=0$$ $$D=(-18)^2-4(27)(-1)=432$$ $$x=\dfrac{18\pm\sqrt{432}}{2(27)}=\dfrac{3\pm2\sqrt{3}}{9}=\dfrac{1}{3}\pm\dfrac{2\sqrt{3}}{9}$$ $$\begin{cases} \dfrac{1}{3}-\dfrac{2\sqrt{3}}{9}\leq x\leq \dfrac{1}{3}+\dfrac{2\sqrt{3}}{9} \\ x>\dfrac{2}{3} \end{cases}$$ $$=>\dfrac{2}{3}<x\leq\dfrac{1}{3}+\dfrac{2\sqrt{3}}{9}$$ $$x\in\bigg(\dfrac{2}{3},\ \dfrac{1}{3}+\dfrac{2\sqrt{3}}{9}\bigg]$$