To analyze the function $f(x) = 2^x + 1$, let's determine its domain and range based on its behavior as an exponential function.

Domain

The function $f(x) = 2^x + 1$ is defined for all real values of $x$. Exponential functions are continuous and defined over the entire set of real numbers.

Domain: $(-\infty, \infty)$

Range

For the range, observe that the function $2^x + 1$ shifts the basic exponential function $2^x$ upward by 1 unit. Since $2^x$ only takes positive values and asymptotically approaches 0 (but never reaches it) as $x$ approaches negative infinity, $f(x)$ will approach 1 from above but will never actually reach 1.

Therefore, the range of $f(x)$ is all values greater than 1.

Range: $(1, \infty)$

Final Answer:

• Domain: $(-\infty, \infty)$
• Range: $(1, \infty)$

Would you like a more detailed explanation of any part of this solution?

Related Questions to Explore:

1. How would the range change if the function were $f(x) = 2^x - 1$?
2. What is the domain and range of the function $g(x) = 3^x + 5$?
3. How would a horizontal shift affect the range of $f(x) = 2^x + 1$?
4. What would be the effect on the domain and range if the function were $f(x) = -2^x + 1$?
5. How can we find the domain and range of a logarithmic function, like $g(x) = \log(x + 1)$?

Tip:

For exponential functions of the form $f(x) = a \cdot b^x + c$, the range typically starts from $c$ (if $a$ and $b$ are positive) and goes to infinity.