Question

Range of the function $f(x) = 3 + 2^x + 4^x$ is

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

Hint:

Since an exponential function $b^x$ can never be zero or negative, the sum $2^x + 4^x$ must be strictly greater than 0 ($2^x + 4^x > 0$). Adding 3 to both sides gives $3 + 2^x + 4^x > 3$. Because it is a strict inequality ($>3$ and not $\ge 3$), use an open parenthesis $(3, \infty)$ instead of a closed bracket.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires finding the complete set of output values (the range) for the exponential function $f(x) = 3 + 2^x + 4^x$ across its natural domain of all real numbers.

Step 2: Key Formula or Approach:
Analyze the behavior of the individual exponential base functions. For any real input $x$, the exponential term $b^x$ (where base $b > 1$) is strictly positive and continuous, approaching $0$ as $x \to -\infty$ and approaching $\infty$ as $x \to \infty$: $$2^x > 0 \quad \text{and} \quad 4^x > 0 \quad \text{for all } x \in \mathbb{R}$$

Step 3: Detailed Explanation:
1. Let's analyze the limits of the function at extreme behaviors of its domain: As $x \to -\infty$: $$2^x \to 0 \quad \text{and} \quad 4^x \to 0$$ Therefore, the function approaches its lower horizontal bound: $$f(x) \to 3 + 0 + 0 = 3$$ Because $2^x$ and $4^x$ can get arbitrarily close to 0 but can never actually equal 0 for any real number $x$, the value $3$ is an open lower boundary ($f(x) > 3$). As $x \to \infty$: Both $2^x \to \infty$ and $4^x \to \infty$. Therefore: $$f(x) \to \infty$$ 2. Since the function is a sum of continuous exponential functions, it is strictly increasing across its entire domain, smoothly covering all values between its lower limit and upper limit.
Thus, the range of the function is strictly values greater than 3, which is written in interval notation as $(3, \infty)$.

Step 4: Final Answer:
The range of the function is $(3, \infty)$, corresponding to option (A).