Answer to Question #263373 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(1 /3x-2)-log₃(x) respectively

1.1. Write down the sets D_f(the domain of f) and D_g(the domain of g)

1.2. Solve the inequality f(x) greater than 2 for x$\in$ D_f

1.3. Solve the inequality g(x) greater or equals to 0 for x$\in$ D_g

Hint: use the change of base formula

Answer 1

1.1 Domain of the function is the set of all values that argument might take. So

$f(x)={\frac 1 {|1-3x|}}$ , then $|1 - 3x| \not=0\implies x \not ={\frac 1 3}$

$D{\scriptscriptstyle f}:x \isin R /$ {${\frac 1 3}$}

$g(x) = \log _{{\frac 1 3}} {{\frac 1 {3x-2}}}-\log _{3} x$ , then $3x-2>0$ and $x>0$

$D{\scriptscriptstyle g}:x \isin ({\frac 2 3};+\infty)$

1.2 ${\frac 1 {|1-3x|}}>2\implies |1-3x|<0.5\implies (1-3x < 0.5)\land (1-3x>-0.5)\implies x \in ({\frac 1 6}; {\frac 1 2})$

Considering the domain, the answer is $x \in ({\frac 1 6}; {\frac 1 2})$ / {${\frac 1 3}$}

1.3 $\log _{{\frac 1 3}} {{\frac 1 {3x-2}}}-\log _{3} x≥0\implies -\log _{3} {{\frac 1 {3x-2}}}-\log _{3} x≥0\implies-(\log _{3} {{\frac 1 {3x-2}}}+\log _{3} x)≥0$

$-(\log _{3} {{\frac 1 {3x-2}}}+\log _{3} x)≥0\implies \log _{3} {{\frac 1 {3x-2}}}+\log _{3} x≤0\implies \log _{3} {{\frac x {3x-2}}} ≤0\implies {\frac x {3x-2}}≤1$

${\frac {x-3x+2} {3x-2}}≤0 \implies {\frac {-2x+2} {3x-2}}≤0\implies ((-2x+2≤0)\land (3x-2>0))\lor((-2x+2 ≥0)\land( 3x-2 < 0))\implies ((x≥1)\land (x>{\frac 2 3}))\lor ((x≤1)\land (x<{\frac 2 3}))\implies (x≥1)\lor (x<{\frac 2 3})$

Considering the domain, the answer is $x\isin [1;+\infty) ∪ (-\infty;{\frac 2 3})$