To find the domain and range of the function $f(x) = \sqrt{2 + 3x - 5x^2}$, follow these steps:

Domain:

The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the square root function, the expression inside the square root must be non-negative (greater than or equal to zero) because the square root of a negative number is not a real number.

1. Set the expression inside the square root to be greater than or equal to zero: $2 + 3x - 5x^2 \geq 0$

2. Rewrite the quadratic inequality: $-5x^2 + 3x + 2 \geq 0$

3. Solve the inequality: First, find the roots of the quadratic equation by setting the expression equal to zero: $-5x^2 + 3x + 2 = 0$ Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -5$, $b = 3$, and $c = 2$.

   $x = \frac{-3 \pm \sqrt{3^2 - 4(-5)(2)}}{2(-5)} = \frac{-3 \pm \sqrt{9 + 40}}{-10} = \frac{-3 \pm \sqrt{49}}{-10} = \frac{-3 \pm 7}{-10}$ So, the roots are: $x = \frac{4}{-10} = -0.4 \quad \text{and} \quad x = \frac{-10}{-10} = 1$

   The roots are $x = -0.4$ and $x = 1$.

4. Determine the intervals where the inequality holds: Test the sign of the expression in each interval determined by the roots $x = -0.4$ and $x = 1$:

   • For $x < -0.4$, the expression is negative.
   • For $-0.4 \leq x \leq 1$, the expression is non-negative.
   • For $x > 1$, the expression is negative.

   Therefore, the domain is: $-0.4 \leq x \leq 1$

Range:

The range of a function is the set of all possible output values (y-values).

1. Evaluate the function at the endpoints of the domain: $f(-0.4) = \sqrt{2 + 3(-0.4) - 5(-0.4)^2} = \sqrt{2 - 1.2 - 0.8} = \sqrt{0} = 0$ $f(1) = \sqrt{2 + 3(1) - 5(1)^2} = \sqrt{2 + 3 - 5} = \sqrt{0} = 0$

2. Check the vertex of the parabola (maximum value within the domain): The vertex occurs at $x = -\frac{b}{2a} = \frac{-3}{-10} = 0.3$. $f(0.3) = \sqrt{2 + 3(0.3) - 5(0.3)^2} = \sqrt{2 + 0.9 - 0.45} = \sqrt{2.45} \approx 1.565$

   Since $f(x)$ is a square root function and has a maximum at $x = 0.3$, the range is: $0 \leq y \leq \sqrt{2.45}$

Final Answer:

• Domain: $-0.4 \leq x \leq 1$
• Range: $0 \leq y \leq \sqrt{2.45}$

Would you like further details or explanations?

Related Questions:

1. How do you solve quadratic inequalities in general?
2. What are the steps to determine the vertex of a quadratic function?
3. How do you find the maximum or minimum value of a quadratic function?
4. What is the significance of the discriminant in finding roots?
5. How can you verify the domain and range of a function graphically?

Tip:

When solving quadratic inequalities, always consider testing intervals around the roots to determine where the function is positive or negative.