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To solve the expression [tex]\( f(x) = \sqrt{x^2 - 3x - 10} \)[/tex], we need to simplify and understand each part of the function. Here's a detailed step-by-step solution:
1. Define the expression inside the square root:
[tex]\( x^2 - 3x - 10 \)[/tex]
2. Simplify the quadratic expression:
The given quadratic expression is already simplified as [tex]\( x^2 - 3x - 10 \)[/tex].
3. Understand the components inside the square root:
[tex]\( x^2 - 3x - 10 \)[/tex] is a quadratic polynomial. The term under the square root determines the domain of the function because the value inside the square root must be non-negative.
4. Determine the domain of [tex]\( f(x) \)[/tex]:
To ensure that the expression under the square root is non-negative (since the square root of a negative number is not a real number), we solve the inequality:
[tex]\( x^2 - 3x - 10 \geq 0 \)[/tex]
5. Find the roots of the quadratic equation:
Solve for [tex]\( x \)[/tex] in the equation [tex]\( x^2 - 3x - 10 = 0 \)[/tex].
To factorize [tex]\( x^2 - 3x - 10 \)[/tex]:
[tex]\[ x^2 - 3x - 10 = (x - 5)(x + 2) = 0 \][/tex]
Therefore, the roots are:
[tex]\[ x = 5 \quad \text{and} \quad x = -2 \][/tex]
6. Determine the intervals:
To find where [tex]\( x^2 - 3x - 10 \)[/tex] is non-negative, consider the intervals determined by the roots:
- [tex]\( (-\infty, -2) \)[/tex]
- [tex]\( (-2, 5) \)[/tex]
- [tex]\( (5, \infty) \)[/tex]
Substitute test points from each interval back into the inequality to determine where the expression is non-negative:
- For [tex]\( x < -2 \)[/tex]: Choose [tex]\( x = -3 \)[/tex]
[tex]\[ (-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8 \quad (positive) \][/tex]
- For [tex]\( -2 < x < 5 \)[/tex]: Choose [tex]\( x = 0 \)[/tex]
[tex]\[ 0^2 - 3(0) - 10 = -10 \quad (negative) \][/tex]
- For [tex]\( x > 5 \)[/tex]: Choose [tex]\( x = 6 \)[/tex]
[tex]\[ 6^2 - 3(6) - 10 = 36 - 18 - 10 = 8 \quad (positive) \][/tex]
Therefore, the expression [tex]\( x^2 - 3x - 10 \geq 0 \)[/tex] is true for:
[tex]\[ x \in (-\infty, -2] \cup [5, \infty) \][/tex]
7. Express the function [tex]\( f(x) \)[/tex] in the domain identified:
Given the domain calculated, [tex]\( f(x) = \sqrt{x^2 - 3x - 10} \)[/tex] is defined for:
[tex]\[ x \in (-\infty, -2] \cup [5, \infty) \][/tex]
8. Conclusion:
The function [tex]\( f(x) = \sqrt{x^2 - 3x - 10} \)[/tex] simplifies to itself, and we have determined the domain as [tex]\( x \in (-\infty, -2] \cup [5, \infty) \)[/tex]. Thus, the function [tex]\( f(x) \)[/tex] can be interpreted as:
[tex]\[ \boxed{\sqrt{x^2 - 3x - 10}} \][/tex]
We have also specified that this function is valid within the domain [tex]\( (-\infty, -2] \cup [5, \infty) \)[/tex].
1. Define the expression inside the square root:
[tex]\( x^2 - 3x - 10 \)[/tex]
2. Simplify the quadratic expression:
The given quadratic expression is already simplified as [tex]\( x^2 - 3x - 10 \)[/tex].
3. Understand the components inside the square root:
[tex]\( x^2 - 3x - 10 \)[/tex] is a quadratic polynomial. The term under the square root determines the domain of the function because the value inside the square root must be non-negative.
4. Determine the domain of [tex]\( f(x) \)[/tex]:
To ensure that the expression under the square root is non-negative (since the square root of a negative number is not a real number), we solve the inequality:
[tex]\( x^2 - 3x - 10 \geq 0 \)[/tex]
5. Find the roots of the quadratic equation:
Solve for [tex]\( x \)[/tex] in the equation [tex]\( x^2 - 3x - 10 = 0 \)[/tex].
To factorize [tex]\( x^2 - 3x - 10 \)[/tex]:
[tex]\[ x^2 - 3x - 10 = (x - 5)(x + 2) = 0 \][/tex]
Therefore, the roots are:
[tex]\[ x = 5 \quad \text{and} \quad x = -2 \][/tex]
6. Determine the intervals:
To find where [tex]\( x^2 - 3x - 10 \)[/tex] is non-negative, consider the intervals determined by the roots:
- [tex]\( (-\infty, -2) \)[/tex]
- [tex]\( (-2, 5) \)[/tex]
- [tex]\( (5, \infty) \)[/tex]
Substitute test points from each interval back into the inequality to determine where the expression is non-negative:
- For [tex]\( x < -2 \)[/tex]: Choose [tex]\( x = -3 \)[/tex]
[tex]\[ (-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8 \quad (positive) \][/tex]
- For [tex]\( -2 < x < 5 \)[/tex]: Choose [tex]\( x = 0 \)[/tex]
[tex]\[ 0^2 - 3(0) - 10 = -10 \quad (negative) \][/tex]
- For [tex]\( x > 5 \)[/tex]: Choose [tex]\( x = 6 \)[/tex]
[tex]\[ 6^2 - 3(6) - 10 = 36 - 18 - 10 = 8 \quad (positive) \][/tex]
Therefore, the expression [tex]\( x^2 - 3x - 10 \geq 0 \)[/tex] is true for:
[tex]\[ x \in (-\infty, -2] \cup [5, \infty) \][/tex]
7. Express the function [tex]\( f(x) \)[/tex] in the domain identified:
Given the domain calculated, [tex]\( f(x) = \sqrt{x^2 - 3x - 10} \)[/tex] is defined for:
[tex]\[ x \in (-\infty, -2] \cup [5, \infty) \][/tex]
8. Conclusion:
The function [tex]\( f(x) = \sqrt{x^2 - 3x - 10} \)[/tex] simplifies to itself, and we have determined the domain as [tex]\( x \in (-\infty, -2] \cup [5, \infty) \)[/tex]. Thus, the function [tex]\( f(x) \)[/tex] can be interpreted as:
[tex]\[ \boxed{\sqrt{x^2 - 3x - 10}} \][/tex]
We have also specified that this function is valid within the domain [tex]\( (-\infty, -2] \cup [5, \infty) \)[/tex].
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