Chapter 3: Problem 41
Find the domain of the given function. $$f(x)=\sqrt{6+x-x^{2}}$$
Short Answer
Expert verified
The domain of \(f(x)\) is \([-2, 3]\).
Step by step solution
01
Identify the Expression Inside the Square Root
The expression inside the square root is \(6 + x - x^2\). For the square root to be defined for real numbers, the expression inside must be non-negative.
02
Set the Expression Greater Than or Equal to Zero
Set up the inequality \(6 + x - x^2 \geq 0\). This will give us the condition for which values of \(x\) make the function defined.
03
Rewrite the Inequality in Standard Quadratic Form
Rearrange the inequality to \(-x^2 + x + 6 \geq 0\). To make it easier to solve, multiply through by -1 (flipping the inequality direction): \(x^2 - x - 6 \leq 0\).
04
Factor the Quadratic Expression
Factor the quadratic expression \(x^2 - x - 6=0\). The factors are \((x - 3)(x + 2) = 0\).
05
Determine the Critical Points
The critical points, or roots, of the equation \((x - 3)(x + 2) = 0\) are \(x = 3\) and \(x = -2\). These points will help us determine the intervals for testing the inequality.
06
Test Intervals Around Critical Points
The critical points divide the number line into intervals: \((-\infty, -2), (-2, 3), (3, \infty)\). Select test points from each interval to check in the inequality \((x - 3)(x + 2) \leq 0\).- For \(x = -3\) in \((-\infty, -2)\), \((-3 - 3)(-3 + 2) > 0\).- For \(x = 0\) in \((-2, 3)\), \((0 - 3)(0 + 2) < 0\).- For \(x = 4\) in \((3, \infty)\), \((4 - 3)(4 + 2) > 0\).The interval \((-2, 3)\) satisfies the inequality.
07
Include Critical Points
Since we have \(\leq\) in our inequality, include \(x = -2\) and \(x = 3\) where the expression equals zero. Check both: - At \(x = -2\), \((-2 - 3)(-2 + 2) = 0\).- At \(x = 3\), \((3 - 3)(3 + 2) = 0\). Both satisfy the inequality.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Square Root Function
The square root function involves finding the non-negative root of a given number or expression. For a square root function, like \(f(x) = \sqrt{6 + x - x^2}\), the expression inside the root must be non-negative. This ensures that the output of the square root is a real number and not an imaginary one.
To determine when \(\sqrt{6 + x - x^2}\) is defined, examine the expression inside the square root: \(6 + x - x^2\). This leads us to set up an inequality, ensuring the expression remains non-negative:
To determine when \(\sqrt{6 + x - x^2}\) is defined, examine the expression inside the square root: \(6 + x - x^2\). This leads us to set up an inequality, ensuring the expression remains non-negative:
- This is the starting point to find the domain of the function.
- By solving \(6 + x - x^2 \geq 0\), we can identify all acceptable \(x\) values.
Quadratic Inequality
A quadratic inequality involves a second-degree polynomial being greater or less than zero. In this context, \(6 + x - x^2 \geq 0\) represents a quadratic inequality essential for determining the domain of our square root function.
Simplifying and rearranging the inequality to a standard quadratic form \(-x^2 + x + 6 \geq 0\), and upon multiplying through by -1, it becomes \(x^2 - x - 6 \leq 0\). The inequality direction changes when multiplying by a negative number.
Simplifying and rearranging the inequality to a standard quadratic form \(-x^2 + x + 6 \geq 0\), and upon multiplying through by -1, it becomes \(x^2 - x - 6 \leq 0\). The inequality direction changes when multiplying by a negative number.
- This form is now easier to solve using methods such as factoring.
- Solutions to the inequality help in revealing intervals where the function is defined.
Factoring Quadratics
Factoring quadratics is a key technique to solve quadratic inequalities or equations. In our example, we are tasked with factoring \(x^2 - x - 6\).
The standard approach is to find two numbers whose product is the constant term (-6) and whose sum is the linear coefficient (-1). For \(x^2 - x - 6\), the factors are \((x - 3)(x + 2) = 0\).
The standard approach is to find two numbers whose product is the constant term (-6) and whose sum is the linear coefficient (-1). For \(x^2 - x - 6\), the factors are \((x - 3)(x + 2) = 0\).
- This forms the foundation to identify "critical points" \(x = 3\) and \(x = -2\), where the polynomial equates to zero.
- These points divide the number line into sections, each of which must be tested in the inequality.