Question

The range of the function $f\left(x\right)=9^x-3^x+1$ is

• A. $(-\infty ,\infty )$
• B. $(-\infty ,0)$
• C. $(0,\infty )$
• D. $[\frac{3}{4},\infty )$

Answer 1

Answer: (D)

$[\frac{3}{4},\infty )$

Let $f\left(x\right)=y=3^{2x}-3^x+1$

$={\left(3^x\right)}^2-3^x+1$

$={(3^x-\frac{1}{2})}^2+1-\frac{1}{4}={(3^x-\frac{1}{2})}^2+\frac{3}{4}$

Now ${(3^x-\frac{1}{2})}^2\ge 0$

$\therefore {(3^x-\frac{1}{2})}^2+\frac{3}{4}\ge \frac{3}{4}$

$\therefore f\left(x\right)\ge \frac{3}{4}$

The required range is $[\frac{3}{4},\infty )$