To find the range of the function $f(x) = \sqrt{2x - 3} + 4$, we need to determine the values that $f(x)$ can take, given the expression for $f(x)$.

Step 1: Understand the domain of the square root function

For the square root function $\sqrt{2x - 3}$ to be valid (i.e., real-valued), the expression under the square root must be non-negative. So, we solve the inequality: $2x - 3 \geq 0$ Solving for $x$: $2x \geq 3$ $x \geq \frac{3}{2}$ Thus, the domain of $f(x)$ is $x \geq \frac{3}{2}$.

Step 2: Analyze the behavior of the function

Now that we know the function is defined for $x \geq \frac{3}{2}$, let's explore the range of the function.

• When $x = \frac{3}{2}$, the expression inside the square root becomes: $\sqrt{2\left(\frac{3}{2}\right) - 3} = \sqrt{3 - 3} = \sqrt{0} = 0$ Therefore, $f\left(\frac{3}{2}\right) = 0 + 4 = 4$.

• As $x \to \infty$, the expression $\sqrt{2x - 3}$ increases without bound, so $f(x)$ also increases without bound. Therefore, as $x \to \infty$, $f(x) \to \infty$.

Step 3: Conclusion

The smallest value of $f(x)$ is 4, which occurs at $x = \frac{3}{2}$, and there is no upper bound, so $f(x)$ can take all values greater than or equal to 4.

Thus, the range of $f(x)$ is: $[4, \infty)$

Would you like further details on this problem or have any questions? Here are some related questions to explore:

1. How does the transformation $+4$ affect the graph of $\sqrt{2x - 3}$?
2. What happens to the range if the square root is replaced with a different function (e.g., quadratic)?
3. Can we determine the domain and range of $f(x)$ without graphing it?
4. How would the range change if the function was $f(x) = \sqrt{2x - 3} - 4$?
5. What is the effect of modifying the coefficient inside the square root (e.g., $f(x) = \sqrt{3x - 3} + 4$)?

Tip: Always consider the domain of the square root function first, as it dictates the valid input values and helps in finding the range.