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: Identify the building blocks.
The function is $f(x) = 3 + 2^x + 4^x$. Both $2^x$ and $4^x$ are exponential terms with base greater than 1, so each is always positive.
Step 2: Note positivity.
For every real $x$, $2^x > 0$ and $4^x > 0$. Hence $2^x + 4^x > 0$ always, so $f(x) > 3$ for all $x$.
Step 3: Check that 3 is never reached.
No real $x$ makes $2^x = 0$ or $4^x = 0$, so $f(x)$ can never actually equal 3. The value 3 is an open lower bound.
Step 4: Examine the left tail.
As $x \to -\infty$, $2^x \to 0$ and $4^x \to 0$, so $f(x) \to 3^{+}$. The graph dips arbitrarily close to 3 but stays above it.
Step 5: Examine the right tail.
As $x \to +\infty$, both $2^x$ and $4^x$ grow without bound, so $f(x) \to \infty$.
Step 6: Combine using continuity.
$f$ is continuous and strictly increasing, so it sweeps through every value strictly greater than 3. The range is $(3, \infty)$.
\[ \boxed{(3, \infty)} \]