Question

In the range of f(x)=6^x+3^x+6^-x+3^-x+2 is a subset of

• A. (6,-∞)
• B. (-2,3)
• C. [-∞,-2]
• D. [6,∞)

Answer 1

The Correct Option is D

To solve for the range of the function \( f(x) = 6^x + 3^x + 6^{-x} + 3^{-x} + 2 \), we must first understand the behavior of each term: \(6^x, 3^x, 6^{-x}, 3^{-x}\).

Step 1: Simplification and Analysis
The expression simplifies to consider positive exponential terms \(6^x\) and \(3^x\) and negative exponential terms \(6^{-x}\) and \(3^{-x}\).
Observe the nature of the function: As \(x\) approaches positive or negative infinity, we examine limits individually.

Step 2: Limit Analysis
 

• As \(x \to \infty\), \(6^x \to \infty\), \(3^x \to \infty\), while \(6^{-x} \to 0\), \(3^{-x} \to 0\). Thus, \(f(x) \to \infty\).
• As \(x \to -\infty\), \(6^x \to 0\), \(3^x \to 0\), while \(6^{-x} \to \infty\), \(3^{-x} \to \infty\). The function again grows large due to the dominance of \(6^{-x}\) and \(3^{-x}\).

Step 3: Minimum Value Calculation
Since the components \(6^x + 6^{-x}\) and \(3^x + 3^{-x}\) are symmetric and reach minimal values of 2 at \(x=0\) due to AM-GM inequality (arithmetic mean-geometric mean inequality), the entire function becomes minimal at this point.

The minimum value of 
\[ f(0) = 6^0 + 3^0 + 6^0 + 3^0 + 2 = 1 + 1 + 1 + 1 + 2 = 6 \]

Step 4: Range Conclusion
Considering the asymptotic behavior and the calculated minimum value, \(f(x)\) ranges from at least 6 upwards. Therefore, the range is \([6, \infty)\), making option [6,∞) the correct answer.