Question

The domain of the function $f\left(x\right)={[9^x+{27}^{\frac{2}{3}(x-2)}-219-3^{2(x-1)}]}^{\frac{1}{4}}$

• A. $[-3,3]$
• B. $[3,\infty )$
• C. $[\frac{5}{2},\infty )$
• D. $[0,1]$

Answer 1

The correct option is C $[\frac{5}{2},\infty )$

For $f\left(x\right)$ to be defined, we must have
$9^x+{27}^{\frac{2}{3}(x-2)}-219-3^{2(x-1)}\ge 0$
$\Rightarrow (3^x{)}^2+3^{2(x-2)}-219-3^{(2x-2)}\ge 0$
$\Rightarrow \left(3^{2x}\right)+\frac{3^{2x}}{81}-219-\frac{3^{2x}}{9}\ge 0$
$\Rightarrow (1+\frac{1}{81}-\frac{1}{9})3^{2x}\ge 219$
$\frac{73}{81}\times 3^{2x}\ge 219$
$\Rightarrow 3^{2x}\ge 3\times 81=3^5$
$\Rightarrow 2x\ge 5$
$\therefore$ Domain is $[\frac{5}{2},\infty ).$