The problem involves finding the domain and range of the function $f(x) = \sqrt{x^2 + 4x - 5}$. Let's analyze this step by step:

a. Domain of $f(x)$

The function $f(x) = \sqrt{x^2 + 4x - 5}$ is defined only when the expression inside the square root, $x^2 + 4x - 5$, is non-negative: $x^2 + 4x - 5 \geq 0$

To solve this inequality, we first find the roots of the quadratic equation $x^2 + 4x - 5 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 4$, and $c = -5$. Substituting: $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-5)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2}$ $x = \frac{-4 \pm 6}{2}$ This gives the roots: $x = 1 \quad \text{and} \quad x = -5$

The quadratic opens upwards (since $a = 1 > 0$), so the inequality $x^2 + 4x - 5 \geq 0$ holds outside the interval between the roots: $x \leq -5 \quad \text{or} \quad x \geq 1$

Thus, the domain of $f(x)$ is: $D(f) = (-\infty, -5] \cup [1, \infty)$

b. Range of $f(x)$

The range of $f(x)$ depends on the possible values of the square root. The minimum value of the quadratic $x^2 + 4x - 5$ occurs at its vertex: $x_{\text{vertex}} = \frac{-b}{2a} = \frac{-4}{2(1)} = -2$ Substitute $x = -2$ into $x^2 + 4x - 5$ to find the minimum value: $(-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9$

Since the quadratic becomes negative ($-9$) at its vertex, $x^2 + 4x - 5$ is not defined for values below 0. Hence, $f(x)$ starts from 0. As $x \to \pm\infty$, the quadratic grows indefinitely, so the square root also grows indefinitely.

Thus, the range of $f(x)$ is: $R(f) = [0, \infty)$

Final Answer:

• Domain: $(-\infty, -5] \cup [1, \infty)$
• Range: $[0, \infty)$

Would you like a detailed graph or further explanation of the steps? Here are some related questions to consider:

1. How can you verify the domain and range graphically for any function?
2. What happens to the domain if the square root is replaced with a cube root?
3. How do you determine the vertex of a quadratic function in standard form?
4. Can the range of this function ever include negative values? Why or why not?
5. How does the sign of the leading coefficient in a quadratic affect its graph?

Tip: Always verify your inequality solutions by testing values in each interval to ensure accuracy.