Answer to Question #284166 in Calculus for mohan

Question

The functions f and g are defined by f(x) =1/(1-3x) and g(x) =log1/3(3x-2)-log3(x) respectively

1. Write down the sets Df (ehe domain of f) and Dg (the domain of g)

2. Solve the inequality f(x) > 2 for x\is in∈ Df

3. Solve the inequality f(x) ≥ 2 for x\is in∈ Dg

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

$$27x^2-18x-1=0$$

$$D=(-18)^2-4(27)(-1)=432$$

$$x=\dfrac{18\pm\sqrt{432}}{2(27)}=\dfrac{3\pm2\sqrt{3}}{9}=\dfrac{1}{3}\pm\dfrac{2\sqrt{3}}{9}$$

$$\begin{cases} \dfrac{1}{3}-\dfrac{2\sqrt{3}}{9}\leq x\leq \dfrac{1}{3}+\dfrac{2\sqrt{3}}{9} \\ x>\dfrac{2}{3} \end{cases}$$

$$=>\dfrac{2}{3}<x\leq\dfrac{1}{3}+\dfrac{2\sqrt{3}}{9}$$

$$x\in\bigg(\dfrac{2}{3},\ \dfrac{1}{3}+\dfrac{2\sqrt{3}}{9}\bigg]$$