Question

The range of$f\left(x\right)=6^x+3^x+6^{-x}+3^{-x}+2$ is a subset of.....

• A. $[-\infty ,-2)$
• B. $(-2,3)$
• C. $(6,\infty )$
• D. $[6,\infty )$

Answer 1

The correct option is D

$[6,\infty )$

The explanation for the correct option.

Compute the range.

Given: $f\left(x\right)=6^x+3^x+6^{-x}+3^{-x}+2$

Add and subtract $4$

$\begin{array}{rcl}f\left(x\right)& =& 6^x+6^{-x}-2+3^x+3^{-x}-2+2+4\\ & =& {\left(\sqrt{6^x}-\sqrt{6^{-x}}\right)}^2+{\left(\sqrt{3^x}-\sqrt{3^{-x}}\right)}^2+6\left[{(a-b)}^2=a^2+b^2-2\mathrm{ab}\right]\end{array}$

We can see that $f\left(x\right)$ is always positive. so the maximum value of $f\left(x\right)$ is $\infty$.

Now, find minimum value of $f\left(x\right)$,

$\begin{array}{rcl}f{\left(x\right)}_{min}& =& 0+0+6\\ & =& 6\end{array}$

Thus, the range of the function $f\left(x\right)$ is $[6,\infty )$.

Hence, option(D) is the correct answer.