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Question Find the domain of \( \mathrm{f}(\mathrm{x}) \). \[ f(x)=\sqrt{-x^{2}+5 x-4} \] Answer Attempt 1 out of 2 Answer Type: Interval watch Vic

Ask by Bowman Pena. in the United States
Jan 24,2025

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The domain of \( f(x) = \sqrt{ -x^2 + 5x - 4 } \) is all real numbers \( x \) such that \( 1 \leq x \leq 4 \). \[ \boxed{[1, 4]} \]

Solution

To find the **domain** of the function \( f(x) = \sqrt{ -x^2 + 5x - 4 } \), we need to determine all real numbers \( x \) for which the expression inside the square root is **non-negative** (since the square root of a negative number is not a real number). ### Step 1: Set the Inside of the Square Root to be Non-Negative \[ -x^2 + 5x - 4 \geq 0 \] ### Step 2: Rearrange the Inequality It's often easier to work with positive leading coefficients. Multiply both sides of the inequality by -1, remembering to **reverse the inequality sign** when multiplying or dividing by a negative number: \[ x^2 - 5x + 4 \leq 0 \] ### Step 3: Factor the Quadratic Expression \[ x^2 - 5x + 4 = (x - 1)(x - 4) \] So the inequality becomes: \[ (x - 1)(x - 4) \leq 0 \] ### Step 4: Find the Critical Points The critical points are the values of \( x \) that make each factor zero: \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] These points divide the real number line into three intervals: 1. \( x < 1 \) 2. \( 1 \leq x \leq 4 \) 3. \( x > 4 \) ### Step 5: Test Each Interval Choose a test point in each interval to determine where the inequality holds. 1. **Interval \( x < 1 \)**: - **Test Point**: \( x = 0 \) - \((0 - 1)(0 - 4) = (-1)(-4) = 4 > 0\) → Doesn't satisfy \( \leq 0 \) 2. **Interval \( 1 \leq x \leq 4 \)**: - **Test Point**: \( x = 2 \) - \((2 - 1)(2 - 4) = (1)(-2) = -2 < 0\) → Satisfies \( \leq 0 \) 3. **Interval \( x > 4 \)**: - **Test Point**: \( x = 5 \) - \((5 - 1)(5 - 4) = (4)(1) = 4 > 0\) → Doesn't satisfy \( \leq 0 \) ### Step 6: Include the Endpoints Check the critical points \( x = 1 \) and \( x = 4 \): - At \( x = 1 \): \[ f(1) = \sqrt{ -1^2 + 5(1) - 4 } = \sqrt{ -1 + 5 - 4 } = \sqrt{0} = 0 \quad \text{(Defined)} \] - At \( x = 4 \): \[ f(4) = \sqrt{ -4^2 + 5(4) - 4 } = \sqrt{ -16 + 20 - 4 } = \sqrt{0} = 0 \quad \text{(Defined)} \] ### Final Answer The function \( f(x) = \sqrt{ -x^2 + 5x - 4 } \) is defined for all real numbers \( x \) in the interval **[1, 4]**. \[ \boxed{\, [\,1,\ 4\,] \,} \]

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The Deep Dive

To find the domain of \( f(x) = \sqrt{-x^2 + 5x - 4} \), we need to ensure that the expression under the square root is non-negative. This translates to solving the inequality: \[ -x^2 + 5x - 4 \geq 0 \] This can be rearranged to: \[ x^2 - 5x + 4 \leq 0 \] Factoring gives: \[ (x - 1)(x - 4) \leq 0 \] Now, we find the critical points by setting the left-hand side to zero: \( x = 1 \) and \( x = 4 \). Testing intervals, we find that the solution is \( 1 \leq x \leq 4 \). Thus, the domain of \( f(x) \) is: \[ [1, 4] \] The square root function is always looking for non-negative values. So, the expression inside has to dance between being greater than or equal to zero to keep the party going! Think of it as only inviting certain numbers that bring positive vibes to your calculation party. And voilà, only numbers from 1 to 4 made the guest list! Want to delve deeper into the methods of inequalities? There’s a treasure trove of resources, including math forums, educational websites like Khan Academy or Coursera, and calculus textbooks that can dive into the beautiful world of real analysis. Exploring these can give you a richer understanding of functions and their domains, along with equations that might pop up during your mathematical journey!

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